Class 11: Binomial Theorem – Exercise 18.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the $11^{th}$ term from the beginning and the $11^{th}$ term from the end in the expansion of $\Big( 2x-$ $\frac{1}{x^2}$ $\Big)^{25}$ .

Answer:

Given expression: $\Big( 2x-$ $\frac{1}{x^2}$ $\Big)^{25}$

Therefore there are $26$ terms in the expansion.

$11^{th}$ term from the end would be $(26-11+1)$ i.e. $16^{th}$ term from the beginning for expression $\Big( 2x-$ $\frac{1}{x^2}$ $\Big)^{25}$

We know that for $(x+a)^n : T_{r+1} = ^{n} C_{r} (x)^{n-r} a^{r}$

Therefore $T_{16} = T_{15+1} = ^{25} C_{15} (2x)^{25-15} \Big(-$ $\frac{1}{x^2}$ $\Big)^{15}$

$= ^{25} C_{15} (2x)^{10} \Big(-$ $\frac{1}{x^{30}}$ $\Big)$

$= - ^{25} C_{15} \Big($ $\frac{2^{10}}{x^{20}}$ $\Big)$

Also $T_{11} = T_{10+1} = ^{25} C_{10} (2x)^{25-10} \Big(-$ $\frac{1}{x^2}$ $\Big)^{10}$

$= ^{25} C_{10} (2x)^{15} \Big($ $\frac{1}{x^{20}}$ $\Big)$

$= ^{25} C_{10} \Big($ $\frac{2^{15}}{x^{5}}$ $\Big)$

$\\$

Question 2: Find the $7^{th}$ term in the expansion $\Big( 3x^2-\frac{1}{x^3} \Big)^{10}$ .

Answer:

Given expression: $\Big( 3x^2- \frac{1}{x^3} \Big)^{10}$

We know that for $(x+a)^n : T_{r+1} = ^{n} C_{r} (x)^{n-r} a^{r}$

$T_{7} = T_{6+1} = ^{10} C_{6} (3x^2)^{10-6} \Big(-$ $\frac{1}{x^3}$ $\Big)^{6}$

$= ^{10} C_{6} \Big($ $\frac{3^4 x^8}{x^{18}}$ $\Big)$

$= ^{10} C_{6} \Big($ $\frac{3^4 }{x^{10}}$ $\Big)$

$=$ $\frac{10 \cdot 9 \cdot 8 \cdot 7 }{4 \cdot 3 \cdot 2 \cdot 1 }$ $\times \Big($ $\frac{3^4}{x^{10}}$ $\Big)$

$=$ $\frac{17010}{x^{10}}$

$\\$

Question 3: Find the $5^{th}$ terms from the end in the expansion of $\Big( 3x-$ $\frac{1}{x^2}$ $\Big)^{10}$ .

Answer:

Given expression: $\Big( 3x-$ $\frac{1}{x^2}$ $\Big)^{10}$

Therefore there are $11$ terms in the expansion.

$5^{th}$ term from the end would be $(11-5+1)$ i.e. $7^{th}$ term from the beginning for expression $\Big( 3x-$ $\frac{1}{x^2}$ $\Big)^{10}$

$T_7 = T_{6+1} = ^{10} C_6 (3x)^{10-6} \Big($ $\frac{-1}{x^2}$ $\Big)^6$

$=$ $\frac{10 \cdot 9 \cdot 8 \cdot 7 }{4 \cdot 3 \cdot 2 \cdot 1 }$ $3^4 \cdot x^4 \cdot$ $\frac{1}{x^{12}}$

$=$ $\frac{17010}{x^8}$

$\\$

Question 4: Find the $8^{th}$ term in the expansion of $( x^{\frac{3}{2}} y^{\frac{1}{2}} - x^{\frac{1}{2}} y^{\frac{3}{2}} )^{10}$ .

Answer:

Given expression: $( x^{\frac{3}{2}} y^{\frac{1}{2}} - x^{\frac{1}{2}} y^{\frac{3}{2}} )^{10}$ .

$T_8 = T_{7+1} = ^{10} C_7$ $( x^{\frac{3}{2}} y^{\frac{1}{2}})^{10-7} ( - x^{\frac{1}{2}} y^{\frac{3}{2}} )^7$

$= - \Big($ $\frac{10 \cdot 9 \cdot 8 }{3 \cdot 2 \cdot 1}$ $\Big)$ $x^{\frac{9}{2}} \cdot y^{\frac{3}{2}} \cdot x^{\frac{7}{2}} \cdot y^{\frac{21}{2}}$

$= - 120 x^8y^{12}$

$\\$

Question 5: Find the $7^{th}$ term in the expansion of $\Big($ $\frac{4x}{5}$ $+$ $\frac{5}{2x}$ $\Big)^8$ .

Answer:

Given expression: $\Big($ $\frac{4x}{5}$ $+$ $\frac{5}{2x}$ $\Big)^8$

$T_7 = T_{6+1} = ^8 C_6 \Big($ $\frac{4x}{5}$ $\Big)^{8-6} \Big($ $\frac{5}{2x}$ $\Big)^{6}$

$=$ $\frac{8 \cdot 7}{2 \cdot 1}$ $\times$ $\frac{4^2 x^2}{5^2}$ $\times$ $\frac{5^6}{2^6 x^6}$

$=$ $\frac{4375}{x^4}$

$\\$

Question 6: Find the $4^{th}$ term in the beginning and the $4^{th}$ term from the end in the expansion $\Big( x +$ $\frac{2}{x}$ $\Big)^9$ .

Answer:

Given expression: $\Big( x +$ $\frac{2}{x}$ $\Big)^9$

Therefore there are $10$ terms in the expansion.

$4^{th}$ term from the end would be $(10-4+1)$ i.e. $7^{th}$ term from the beginning for expression $\Big( x +$ $\frac{2}{x}$ $\Big)^9$

$T_7 = T_{6+1} =^9 C_6 ( x)^{9-6} \Big($ $\frac{2}{x}$ $\Big)^6$

$=$ $\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}$ $\cdot x^3\cdot$ $\frac{2^6}{x^6}$

$=$ $\frac{5376}{x^3}$

$T_4 = T_{3+1} =^9 C_3 ( x)^{9-3} \Big($ $\frac{2}{x}$ $\Big)^3$

$=$ $\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}$ $\cdot x^6\cdot$ $\frac{2^3}{x^3}$

$= 672x^3$

$\\$

Question 7: Find the $4^{th}$ term from the end in the expansion of $\Big($ $\frac{4x}{5}$ $-$ $\frac{5}{2x}$ $\Big)^9$ .

Answer:

Given expression: $\Big($ $\frac{4x}{5}$ $-$ $\frac{5}{2x}$ $\Big)^9$

Therefore there are $10$ terms in the expansion.

$4^{th}$ term from the end would be $(10-4+1)$ i.e. $7^{th}$ term from the beginning for expression $\Big($ $\frac{4x}{5}$ $-$ $\frac{5}{2x}$ $\Big)^9$

$T_7 = T_{6+1} = ^9 C_6 \Big($ $\frac{4x}{5}$ $\Big)^{9-6} \Big($ $\frac{-5}{2x}$ $\Big)^{6}$

$=$ $\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}$ $\cdot$ $\frac{4^3x^3}{5^3}$ $\cdot$ $\frac{5^6}{2^6x^6}$

$=$ $\frac{10500}{x^3}$

$\\$

Question 8: Find the $7^{th}$ term from the end in the expansion of $\Big( 2x^2 -$ $\frac{3}{2x}$ $\Big)^8$ .

Answer:

Given expression: $\Big( 2x^2 -$ $\frac{3}{2x}$ $\Big)^8$

Therefore there are $10$ terms in the expansion.

$7^{th}$ term from the end would be $(9-7+1)$ i.e. $3^{rd}$ term from the beginning for expression $\Big( 2x^2 -$ $\frac{3}{2x}$ $\Big)^8$

$T_3 = T_{2+1} = ^8 C_2 (2x^2)^{8-2} \Big($ $\frac{-3}{2x}$ $\Big)^2$

$=$ $\frac{8 \cdot 7 }{2 \cdot 1}$ $\cdot 2^6 x^{12} \cdot \Big($ $\frac{3^2}{2^2x^2}$ $\Big)$

$= 4032 x^{10}$

$\\$

Question 9: Find the coefficient of:

i) $x^{10}$ in the expansion of $\Big( 2x^2 -$ $\frac{1}{x}$ $\Big)^{20}$

ii) $x^{7}$ in the expansion of $\Big( x -$ $\frac{1}{x^2}$ $\Big)^{40}$

iii) $x^{-15}$ in the expansion of $\Big( 3x^2 -$ $\frac{a}{3x^3}$ $\Big)^{10}$

iv) $x^{9}$ in the expansion of $\Big( x^2 -$ $\frac{1}{3x}$ $\Big)^{9}$

v) $x^{m}$ in the expansion of $\Big( x +$ $\frac{1}{x}$ $\Big)^{n}$

vi) $x$ in the expansion of $( 1- 2x^3 + 3x^5) \Big( 1+$ $\frac{1}{x}$ $\Big)^{8}$

vii) $a^5b^7$ in the expansion of $(a-2b)^{12}$

viii) $x$ in the expansion of $( 1 - 3x + 7x^2)( 1 - x)^{16}$

Answer:

i) The given expression: $\Big( 2x^2 -$ $\frac{1}{x}$ $\Big)^{20}$ .

Let $(r+1)^{th}$ term has $x^{10}$

$T_{r+1} = ^{20} C_{r} \cdot (2x^2)^{20-r} \cdot \Big($ $\frac{-1}{x}$ $\Big)^r$

$= ^{20} C_{r} \cdot 2^{20-r} \cdot x^{40-2r} \cdot \Big($ $\frac{-1^r}{x^r}$ $\Big)$

$= (-1)^r \cdot {^{20} C_{r}} \cdot 2^{20-r} \cdot x^{40-3r}$

Therefore $40-3r= 10 \Rightarrow r = 10$

Therefore coefficient of $x^{10} = (-1)^{10} \cdot {^{20} C_{10}} \cdot 2^{20-10} = ^{20} C_{10} \cdot 2^{10}$

$\\$

ii) The given expression: $\Big( x -$ $\frac{1}{x^2}$ $\Big)^{40}$ .

Let $(r+1)^{th}$ term has $x^{7}$

$T_{r+1} = ^{40} C_{r} \cdot (x)^{40-r} \cdot \Big($ $\frac{-1}{x^2}$ $\Big)^r$

$= (-1)^r {^{40} C_{r}} \cdot \Big($ $\frac{x^{40-r}}{x^{2r}}$ $\Big)$

$= (-1)^r \cdot {^{40} C_{r}} \cdot x^{40-3r}$

Therefore $40-3r= 7 \Rightarrow r = 11$

Therefore coefficient of $x^{7} = (-1)^{11} \cdot {^{40} C_{11}} = - ^{40} C_{11}$

$\\$

iii) The given expression: $\Big( 3x^2 -$ $\frac{a}{3x^3}$ $\Big)^{10}$

Let $(r+1)^{th}$ term has $x^{-15}$

$T_{r+1} = ^{10} C_{r} \cdot (3x^2)^{10-r} \cdot \Big($ $\frac{-a}{3x^3}$ $\Big)^r$

$= (-1)^r {^{10} C_{r}} \cdot 3^{10-r} x^{20-2r} \Big($ $\frac{a^r}{3^r x^{3r}}$ $\Big)$

$= (-1)^r \cdot {^{10} C_{r}} 3^{10-2r} a^r \cdot x^{20-5r}$

Therefore $20-5r= -15 \Rightarrow r = 7$

Therefore coefficient of $x^{7} = (-1)^{7} \cdot {^{10} C_{7}} 3^{10-14} a^7 = - ^{10} C_{7}$ $\frac{a^7}{3^4}$ $=$ $\frac{-40}{9}$ $a^7$

$\\$

iv) The given expression: $\Big( x^2 -$ $\frac{1}{3x}$ $\Big)^{9}$ .

Let $(r+1)^{th}$ term has $x^{9}$

$T_{r+1} = ^{9} C_{r} \cdot (x^2)^{9-r} \cdot \Big($ $\frac{-1}{3x}$ $\Big)^r$

$= (-1)^r \cdot {^{9} C_{r}} \cdot x^{18-2r} \Big($ $\frac{1}{3^r x^{r}}$ $\Big)$

$= (-1)^r \cdot {^{9} C_{r}} \cdot$ $\frac{1}{3^r}$ $\cdot x^{18-3r}$

Therefore $18-3r= 9 \Rightarrow r = 3$

Therefore coefficient of $x^{9} = (-1)^{3} \cdot {^{9} C_{3}} \cdot$ $\frac{1}{3^3}$ $= (-1) \cdot$ $\frac{9 \cdot 8 \cdot 7 }{3 \cdot 2 \cdot 1 }$ $\cdot$ $\frac{1}{3^3}$ $=$ $\frac{-28}{9}$

$\\$

v) The given expression: $\Big( x +$ $\frac{1}{x}$ $\Big)^{n}$ .

Let $(r+1)^{th}$ term has $x^{m}$

$T_{r+1} = ^{n} C_{r} \cdot (x)^{n-r} \cdot \Big($ $\frac{1}{x}$ $\Big)^r$

$= {^{n} C_{r}} \cdot x^{n-2r}$

Therefore $n-2r= m \Rightarrow r =$ $\frac{n-m}{2}$

Therefore coefficient of $x^{m} = ^{n} C_{\frac{n-m}{2}} =$ $\frac{n!}{(\frac{n-m}{2})! (\frac{n+m}{2})!}$

$\\$

vi) Given expression: $( 1- 2x^3 + 3x^5) \Big( 1+$ $\frac{1}{x}$ $\Big)^{8}$

$= ( 1- 2x^3 + 3x^5) \Big[ ^8 C_0 \Big( \frac{1}{x} \Big)^0 + ^8 C_1 \Big( \frac{1}{x} \Big)^1 +^8 C_2 \Big( \frac{1}{x} \Big)^2 +^8 C_3 \Big( \frac{1}{x} \Big)^3 +^8 C_4 \Big( \frac{1}{x} \Big)^4 \\ \\ { \hspace{5.0cm}+^8 C_5 \Big( \frac{1}{x} \Big)^5 +^8 C_6 \Big( \frac{1}{x} \Big)^6 +^8 C_7 \Big( \frac{1}{x} \Big)^7 + ^8 C_8 \Big( \frac{1}{x} \Big)^8} \Big]$

$= ( 1- 2x^3 + 3x^5) \Big[ ^8 C_0 + ^8 C_1 \Big( \frac{1}{x} \Big) +^8 C_2 \Big( \frac{1}{x^2} \Big) +^8 C_3 \Big( \frac{1}{x^3} \Big) +^8 C_4 \Big( \frac{1}{x^4} \Big) \\ \\ { \hspace{5.0cm}+^8 C_5 \Big( \frac{1}{x^5} \Big) +^8 C_6 \Big( \frac{1}{x^6} \Big) +^8 C_7 \Big( \frac{1}{x^7} \Big) + ^8 C_8 \Big( \frac{1}{x^8} \Big)} \Big]$

Therefore terms with $x = -2x^3 \cdot ^8 C_2 \cdot$ $\frac{1}{x^2}$ $+ 3x^5 \cdot ^8 C_4 \cdot$ $\frac{1}{x^4}$

$= -2 \cdot ^8 C_2 \cdot x + 3 \cdot ^8 C_4 \cdot x$

$= \Big( -2 \cdot$ $\frac{8 \cdot 7}{2 \cdot 1}$ $+ 3 \cdot$ $\frac{8 \cdot7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}$ $\Big) x$

$= (-56+210) x$

$= 154x$

Hence the coefficient of the term with $x = 154$

$\\$

vii) Given expression: $(a-2b)^{12}$

$= ^{12} C_0 (a)^{12} (-2b)^0 + ^{12} C_1 (a)^{11} (-2b)^1 + ^{12} C_2 (a)^{10} (-2b)^2 \\ \\ { \hspace{3.0cm} + ^{12} C_3 (a)^{9} (-2b)^3 + ^{12} C_4 (a)^{8} (-2b)^4 + ^{12} C_5 (a)^{7} (-2b)^5} \\ \\ {\hspace{3.0cm} + ^{12} C_6 (a)^{6} (-2b)^6 + ^{12} C_7 (a)^{5} (-2b)^7 + \ldots }$

Therefore coefficient of $a^5b^7 = ^{12} C_7 (-2)^7 = -$ $\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }$ $\cdot 2^7 = - 101376$

$\\$

viii) Given expression: $( 1 - 3x + 7x^2)( 1 - x)^{16}$

$= ( 1 - 3x + 7x^2) [ ^{16} C_0 (-x)^0 + ^{16} C_1 (-x)^1 + \ldots ]$

$= ( 1 - 3x + 7x^2) [ 1 + 16(-x) + \ldots ]$

Therefore coefficient of $x = -3 + ( -16) = - 19$

$\\$

Question 10: Which term in the expansion of $\Big\{ \Big( \frac{x}{\sqrt{y}} \Big)^{\frac{1}{3}} + \Big( \frac{y}{x^{\frac{1}{3}}} \Big)^{\frac{1}{2}} \Big\}^{21}$ contains $x$ and $y$ to one and the same power?

Answer:

Given expression: $\Big\{ \Big( \frac{x}{\sqrt{y}} \Big)^{\frac{1}{3}} + \Big( \frac{y}{x^{\frac{1}{3}}} \Big)^{\frac{1}{2}} \Big\}^{21}$

$T_{r+1} = ^{21} C_r \cdot$ $\Big[ \Big( \frac{x}{\sqrt{y}} \Big)^{\frac{1}{3}} \Big]^{21-r} \cdot \Big[ \Big( \frac{y}{x^{\frac{1}{3}}} \Big)^{\frac{1}{2}} \Big]^{r}$

$= ^{21} C_r \cdot$ $\Big[ \frac{x^{\frac{21-r}{3}}}{y^{\frac{21-r}{6}}} \Big] \cdot \Big[ \frac{y^{\frac{r}{2}}}{x^{\frac{r}{6}}} \Big]$

$= ^{21} C_r \cdot$ $\frac{ x^{\frac{21-r}{3} - \frac{r}{6}}}{y^{\frac{21-r}{6} - \frac{r}{2}}}$

$= ^{21} C_r \cdot$ $\frac{x^{\frac{42-3r}{6}}}{y^{\frac{21-4r}{6}}}$

$= ^{21} C_r \cdot$ $x^{7 - \frac{r}{2}} \cdot y^{ \frac{2}{3} r - \frac{7}{2}}$

Therefore $7 -$ $\frac{r}{2} = \frac{2}{3} r - \frac{7}{2}$

$\frac{21}{2} = \frac{7r}{6}$ $\Rightarrow r = 9$

Therefore the required term $= 10^{th}$ term.

$\\$

Question 11: Does the expansion of $\Big( 2x^2 -$ $\frac{1}{x}$ $\Big)^{20}$ contain any term involving $x^{9}$ .

Answer:

General expression: $\Big( 2x^2 -$ $\frac{1}{x}$ $\Big)^{20}$

Suppose $x^9$ occurs in the given expression at the $( r+1)^{th}$ term.

$T_{r+1}= ^{20} C_r \cdot (2x^2)^{20-r} \cdot \Big($ $\frac{-1}{x}$ $\Big)^r$

$= (-1)^r \cdot {^{20} C_r} \cdot 2^{20-r} \cdot x^{40-2r-r}$

Fr the term to contain $x^9$ ,

$40-3r = 9 \Rightarrow r =$ $\frac{31}{3}$

Since $r$ can only be an integer, this is not possible. Hence there is no term in the expansion of $\Big( 2x^2 -$ $\frac{1}{x}$ $\Big)^{20}$ that contains $x^9$ .

$\\$

Question 12: Show that the expansion of $\Big( x^2 +$ $\frac{1}{x}$ $\Big)^{12}$ does not contain any term involving $x^{-1}$ .

Answer:

Given expression: $\Big( x^2 +$ $\frac{1}{x}$ $\Big)^{12}$

Suppose $x^{-1}$ occurs in the given expression at the $( r+1)^{th}$ term.

$T_{r+1}= ^{12} C_r \cdot (x^2)^{12-r} \cdot \Big($ $\frac{1}{x}$ $\Big)^r$

$= ^{12} C_r \cdot x^{24-2r-r}$

For the term to contain $x^{-1}$ ,

$24-3r = -1 \Rightarrow r =$ $\frac{25}{3}$

Since $r$ can only be an integer, this is not possible. Hence there is no term in the expansion of $\Big( x^2 +$ $\frac{1}{x}$ $\Big)^{12}$ .

$\\$

Question 13: Find the middle term in the expansion of:

i) $\Big($ $\frac{2x}{3}$ $-$ $\frac{3}{2x}$ $\Big)^{20}$ ii) $\Big($ $\frac{a}{x}$ $+ bx \Big)^{12}$ iii) $\Big( x^2 -$ $\frac{2}{x}$ $\Big)^{10}$ iv) $\Big($ $\frac{x}{a}$ $-$ $\frac{a}{x}$ $\Big)^{10}$

Answer:

i) $\Big($ $\frac{2x}{3}$ $-$ $\frac{3}{2x}$ $\Big)^{20}$

Here $n = 20$ is even. Therefore $\Big($ $\frac{20}{2}$ $+1\Big)$ i.e. $11^{th}$ term is the middle term.

$T_{11} = T_{10+1} = ^{20} C_{10} \cdot \Big($ $\frac{2}{3}$ $x \Big)^{20-10} \cdot \Big($ $\frac{-3}{2x}$ $\Big)^{10}$

$= ^{20} C_{10} \cdot \Big($ $\frac{2}{3}$ $\Big)^{10} x^{10} \cdot \Big($ $\frac{3}{2}$ $\Big)^{10}$ $\frac{1}{x^{10}}$

$= ^{20} C_{10}$

ii) $\Big($ $\frac{a}{x}$ $+ bx \Big)^{12}$

Here $n = 12$ is even. Therefore $\Big($ $\frac{12}{2}$ $+1\Big)$ i.e. $7^{th}$ term is the middle term.

$T_{7} = T_{6+1} = ^{12} C_{6} \cdot \Big($ $\frac{a}{x}$ $\Big)^{12-6} \cdot (bx)^6$

$= ^{12} C_{6} a^6b^6$

$= 924 a^6b^6$

iii) $\Big( x^2 -$ $\frac{2}{x}$ $\Big)^{10}$

Here $n = 10$ is even. Therefore $\Big($ $\frac{10}{2}$ $+1\Big)$ i.e. $6^{th}$ term is the middle term.

$T_{6} = T_{5+1} = ^{10} C_{5} \cdot (x^2)^{10-5} \cdot \Big($ $\frac{-2}{x}$ $\Big)^{5}$

$= - ^{10} C_{5} \cdot x^{10} \cdot \Big($ $\frac{2^5}{x^5}$ $\Big)$

$= -$ $\frac{ 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$ $\times 2^5 \times x^5$

$= -8064 x^5$

iv) $\Big($ $\frac{x}{a}$ $-$ $\frac{a}{x}$ $\Big)^{10}$

Here $n = 10$ is even. Therefore $\Big($ $\frac{10}{2}$ $+1\Big)$ i.e. $6^{th}$ term is the middle term.

$T_{6} = T_{5+1} = ^{10} C_{5} \cdot \Big($ $\frac{x}{a}$ $\Big)^{10-5} \cdot \Big($ $\frac{-a}{x}$ $\Big)^{5}$

$= - ^{10} C_{5} \cdot \Big($ $\frac{x^5}{a^5}$ $\Big) \cdot \Big($ $\frac{a^5}{x^5}$ $\Big)$

$= - ^{10} C_{5} = -252$

$\\$

Question 14: Find the middle term in the expansion of:

i) $\Big( 3x -$ $\frac{x^3}{6}$ $\Big)^{9}$ ii) $\Big( 2x^2-$ $\frac{1}{x}$ $\Big)^{7}$ iii) $\Big( 3x -$ $\frac{2}{x^2}$ $\Big)^{15}$ iv) $\Big( x^4 -$ $\frac{1}{x^3}$ $\Big)^{11}$

Answer:

i) Given expression: $\Big( 3x -$ $\frac{x^3}{6}$ $\Big)^{9}$

Here $n = 9$ is odd. Therefore $\Big($ $\frac{9+1}{2}$ $\Big)$ and $\Big($ $\frac{9+1}{2}$ $+ 1 \Big)$ i.e. $5^{th}$ and $6^{th}$ terms are the middle term of the given expression.

$T_{5} = T_{4+1} = ^{9} C_{4} \cdot (3x)^{9-4} \cdot \Big($ $\frac{-x^3}{6}$ $\Big)^4$

$= ^{9} C_{4} \cdot 3^5 \cdot x^5 \cdot$ $\frac{x^{12}}{6^4}$

$=$ $\frac{9 \cdot 8 \cdot 7 \cdot 6}{ 4 \cdot 3 \cdot 2 \cdot 1}$ $\times$ $\frac{3^5}{6^4}$ $\times x^{17}$

$=$ $\frac{189}{8}$ $x^{17}$

$T_{6} = T_{5+1} = ^{9} C_{5} \cdot (3x)^{9-5} \cdot \Big($ $\frac{-x^3}{6}$ $\Big)^5$

$= - ^{9} C_{5} \cdot 3^4 \cdot x^4 \cdot$ $\frac{x^{15}}{6^5}$

$= -$ $\frac{9 \cdot 8 \cdot 7 \cdot 6}{ 4 \cdot 3 \cdot 2 \cdot 1}$ $\times$ $\frac{3^4}{6^5}$ $\times x^{19}$

$= -$ $\frac{21}{16}$ $x^{19}$

ii) Given expression: $\Big( 2x^2-$ $\frac{1}{x}$ $\Big)^{7}$

Here $n = 7$ is odd. Therefore $\Big($ $\frac{7+1}{2}$ $\Big)$ and $\Big($ $\frac{7+1}{2}$ $+ 1 \Big)$ i.e. $4^{th}$ and $5^{th}$ terms are the middle term of the given expression.

$T_{4} = T_{3+1} = ^{7} C_{3} \cdot (2x^2)^{7-3} \cdot \Big($ $\frac{-1}{x}$ $\Big)^3$

$= - ^{7} C_{3} \cdot 2^4 \cdot x^8 \cdot$ $\frac{1}{x^3}$

$= -$ $\frac{7 \cdot 6 \cdot 5}{ 3 \cdot 2 \cdot 1}$ $\times 2^4 \times x^5$

$= - 560x^5$

$T_{5} = T_{4+1} = ^{7} C_{4} \cdot (2x^2)^{7-4} \cdot \Big($ $\frac{-1}{x}$ $\Big)^4$

$= ^{7} C_{4} \cdot 2^3 \cdot x^6 \cdot$ $\frac{1}{x^4}$

$=$ $\frac{7 \cdot 6 \cdot 5}{ 3 \cdot 2 \cdot 1}$ $\times 2^3 \times x^2$

$= 280x^2$

iii) Given expression: $\Big( 3x -$ $\frac{2}{x^2}$ $\Big)^{15}$

Here $n = 15$ is odd. Therefore $\Big($ $\frac{15+1}{2}$ $\Big)$ and $\Big($ $\frac{15+1}{2}$ $+ 1 \Big)$ i.e. $8^{th}$ and $9^{th}$ terms are the middle term of the given expression.

$T_{8} = T_{7+1} = ^{15} C_{7} \cdot (3x)^{15-7} \cdot \Big($ $\frac{-2}{x^2}$ $\Big)^7$

$= - ^{15} C_{7} \cdot 3^8 \cdot x^8 \cdot$ $\frac{2^7}{x^{14}}$

$= -$ $\frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 }{ 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$ $\times 3^8 \times 2^7 \times$ $\frac{1}{x^6}$

$= -$ $\frac{6435 \times 3^8 \times 2^7}{x^6}$

$T_{9} = T_{8+1} = ^{15} C_{8} \cdot (3x)^{15-8} \cdot \Big($ $\frac{-2}{x^2}$ $\Big)^8$

$= ^{15} C_{8} \cdot 3^7 \cdot x^7 \cdot$ $\frac{2^8}{x^{16}}$

$=$ $\frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 }{ 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$ $\times 3^7 \times 2^8 \times$ $\frac{1}{x^9}$

$=$ $\frac{6435 \times 3^7 \times 2^8}{x^9}$

iv) Given expression: $\Big( x^4 -$ $\frac{1}{x^3}$ $\Big)^{11}$

Here $n = 11$ is odd. Therefore $\Big($ $\frac{11+1}{2}$ $\Big)$ and $\Big($ $\frac{11+1}{2}$ $+ 1 \Big)$ i.e. $6^{th}$ and $7^{th}$ terms are the middle term of the given expression.

$T_{6} = T_{5+1} = ^{11} C_{5} \cdot (x^4)^{11-5} \cdot \Big($ $\frac{-1}{x^3}$ $\Big)^5$

$= - ^{11} C_{5} \cdot x^{24} \cdot$ $\frac{1}{x^{15}}$

$= -$ $\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{ 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$ $\times x^9$

$= - 462x^9$

$T_{7} = T_{6+1} = ^{11} C_{6} \cdot (x^4)^{11-6} \cdot \Big($ $\frac{-1}{x^3}$ $\Big)^6$

$= ^{11} C_{6} \cdot x^{20} \cdot$ $\frac{1}{x^{18}}$

$=$ $\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{ 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$ $x^2$

$= 462x^2$

$\\$

Question 15: Find the middle term in the expansion of:

i) $\Big( x -$ $\frac{1}{x}$ $\Big)^{10}$ ii) $(1-2x+x^2)^n$ iii) $(1+3x+3x^2+x^3)^{2n}$

iv) $\Big( 2x -$ $\frac{x^2}{4}$ $\Big)^{9}$ v) $\Big( x -$ $\frac{1}{x}$ $\Big)^{2n+1}$ vi) $\Big($ $\frac{x}{3}$ $+9y \Big)^{10}$

vii) $\Big( 3 -$ $\frac{x^3}{6}$ $\Big)^{7}$ viii) $\Big( 2ax -$ $\frac{b}{x^2}$ $\Big)^{12}$ ix) $\Big($ $\frac{p}{x}$ $+$ $\frac{x}{p}$ $\Big)^{9}$ x) $\Big($ $\frac{x}{a}$ $-$ $\frac{a}{x}$ $\Big)^{10}$

Answer:

i) Given expression: $\Big( x -$ $\frac{1}{x}$ $\Big)^{10}$

Here $n = 10$ is even. Therefore $\Big($ $\frac{10}{2}$ $+1\Big)$ i.e. $6^{th}$ term is the middle term.

$T_6 = T_{5+1} = ^{10} C_5 \cdot (x)^{10-5} \cdot \Big($ $\frac{-1}{x}$ $\Big)^5$

$= - ^{10} C_5 \cdot$ $\frac{x^5}{x^5}$

$= -$ $\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 }{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

$= - 252$

ii) Given expression: $(1-2x+x^2)^n = (1-x)^{2n}$

Here $2n$ is even. Therefore $\Big($ $\frac{2n}{2}$ $+1\Big)$ i.e. $(n+1)^{th}$ term is the middle term.

$T_{n+1} = ^{2n} C_n \cdot (1)^{2n-n} \cdot ( -x )^n$

$= (-1)^n \cdot {^{2n} C_n} \cdot x^n$

$= (-1)^n$ $\frac{(2n)!}{(n!)^2}$ $x^n$

iii) Given expression: $(1+3x+3x^2+x^3)^{2n} = {(1+x)^3}^{2n} = (1+x)^{6n}$

Here $6n$ is even. Therefore $\Big($ $\frac{6n}{2}$ $+1\Big)$ i.e. $(3n+1)^{th}$ term is the middle term.

$T_{3n+1} = ^{6n} C_{3n} \cdot (1)^{6n-3n} \cdot (x)^{3n}$

$= ^{6n} C_{3n} \cdot (x)^{3n}$

$=$ $\frac{(6n)!}{[(3n)!]^2}$ $(x)^{3n}$

iv) Given expression: $\Big( 2x -$ $\frac{x^2}{4}$ $\Big)^{9}$

Here $n = 9$ is odd. Therefore $\Big($ $\frac{9+1}{2}$ $\Big)$ and $\Big($ $\frac{9+1}{2}$ $+ 1 \Big)$ i.e. $5^{th}$ and $6^{th}$ terms are the middle term of the given expression.

$T_5 = T_{4+1} = ^{9} C_4 \cdot (2x)^{9-4} \cdot \Big($ $\frac{-x^2}{4}$ $\Big)^4$

$= ^{9} C_4 \cdot 2^5 x^5 \cdot$ $\frac{x^8}{4^4}$

$=$ $\frac{9 \cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1}$ $\times$ $\frac{2^5}{4^4}$ $x^{13}$

$=$ $\frac{63}{4}$ $x^{13}$

$T_6 = T_{5+1} = ^{9} C_5 \cdot (2x)^{9-5} \cdot \Big($ $\frac{-x^2}{4}$ $\Big)^5$

$= - ^{9} C_5 \cdot 2^4 x^4 \cdot$ $\frac{x^{10}}{4^5}$

$= -$ $\frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2 \cdot 1}$ $\times$ $\frac{2^4}{4^5}$ $x^{14}$

$= -$ $\frac{63}{32}$ $x^{14}$

v) Given expression: $\Big( x -$ $\frac{1}{x}$ $\Big)^{2n+1}$

Here $n = (2n+1)$ is odd. Therefore $\Big($ $\frac{2n+1+1}{2}$ $\Big)$ and $\Big($ $\frac{2n+1+1}{2}$ $+ 1 \Big)$ i.e. $(n+1)^{th}$ and $(n+2)^{th}$ terms are the middle term of the given expression.

$T_{n+1} = ^{2n+1} C_n \cdot (x)^{2n+1-n} \cdot \Big($ $\frac{-1}{x}$ $\Big)^n$

$= {^{2n+1} C_n} \cdot x^{n+1} \cdot$ $\frac{(-1)^n}{x^n}$

$= (-1)^n \cdot {^{2n+1} C_n} x$

$T_{n+2} = T_{n+1+1} = ^{2n+1} C_{n+1} \cdot (x)^{2n+1-(n+1)} \cdot \Big($ $\frac{-1}{x}$ $\Big)^{n+1}$

$= (-1)^{n+1} \cdot {^{2n+1} C_n} \cdot x^{n} \cdot$ $\frac{1}{x^{n+1}}$

$= (-1)^{n+1} \cdot {^{2n+1} C_n} \cdot$ $\frac{1}{x}$

vi) Given expression: $\Big( \frac{x}{3}+9y \Big)^{10}$

Here $n = 10$ is even. Therefore $\Big($ $\frac{10}{2}$ $+1\Big)$ i.e. $6^{th}$ term is the middle term.

$T_6 = T_{5+1} = ^{10} C_5 \cdot \Big($ $\frac{x}{3}$ $\Big)^{10-5} \cdot (9y)^5$

$= ^{10} C_5 \cdot$ $\frac{x^5}{3^5}$ $\cdot 9^5 \cdot y^5$

$= -$ $\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 }{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{x^5}{3^5}$ $\cdot 9^5 \cdot y^5$

$= 61236 x^5y^5$

vii) Given expression: $\Big( 3 -$ $\frac{x^3}{6}$ $\Big)^{7}$

Here $n = 7$ is odd. Therefore $\Big($ $\frac{7+1}{2}$ $\Big)$ and $\Big($ $\frac{7+1}{2}$ $+ 1 \Big)$ i.e. $4^{th}$ and $5^{th}$ terms are the middle term of the given expression.

$T_4 = T_{3+1} = ^{7} C_7 \cdot (3)^{7-3} \cdot \Big($ $\frac{-x^3}{6}$ $\Big)^3$

$= (-1) {^{7} C_3} \cdot 3^4 \cdot$ $\frac{x^9}{6^3}$

$= -$ $\frac{105}{8}$ $x^{9}$

$T_5 = T_{4+1} = ^{7} C_4 \cdot (3)^{7-4} \cdot \Big($ $\frac{-x^3}{6}$ $\Big)^4$

$= {^{7} C_4} \cdot 3^3 \cdot$ $\frac{x^{12}}{6^4}$

$=$ $\frac{35}{48}$ $x^{12}$

viii) Given expression: $\Big( 2ax -$ $\frac{b}{x^2}$ $\Big)^{12}$

Here $n = 12$ is even. Therefore $\Big($ $\frac{12}{2}$ $+1\Big)$ i.e. $7^{th}$ term is the middle term.

$T_7 = T_{6+1} = ^{12} C_6 \cdot (2ax)^{12-6} \cdot \Big($ $\frac{-b}{x^2}$ $\Big)^6$

$= ^{12} C_6 \cdot 2^6 a^6 x^6 \cdot$ $\frac{b^6}{x^{12}}$

$= 59136 \Big($ $\frac{a^6 b^6 }{x^6}$ $\Big)$

ix) Given expression: $\Big($ $\frac{p}{x}$ $+$ $\frac{x}{p}$ $\Big)^{9}$

Here $n = 9$ is odd. Therefore $\Big($ $\frac{9+1}{2}$ $\Big)$ and $\Big($ $\frac{9+1}{2}$ $+ 1 \Big)$ i.e. $5^{th}$ and $6^{th}$ terms are the middle term of the given expression.

$T_5 = T_{4+1} = ^{9} C_4 \cdot \Big($ $\frac{p}{x}$ $\Big)^{9-4} \cdot \Big($ $\frac{x}{p}$ $\Big)^4$

$= {^{9} C_4} \cdot \Big($ $\frac{p}{x}$ $\Big)^5 \cdot \Big($ $\frac{x}{p}$ $\Big)^4$

$= {^{9} C_4} \Big($ $\frac{p}{x}$ $\Big)$

$= 126 \Big($ $\frac{p}{x}$ $\Big)$

$T_6 = T_{5+1} = ^{9} C_5 \cdot \Big($ $\frac{p}{x}$ $\Big)^{9-5} \cdot \Big($ $\frac{x}{p}$ $\Big)^5$

$= {^{9} C_5} \cdot \Big($ $\frac{p}{x}$ $\Big)^4 \cdot \Big($ $\frac{x}{p}$ $\Big)^5$

$= {^{9} C_4} \cdot \Big($ $\frac{x}{p}$ $\Big)$

$= 126 \Big($ $\frac{x}{p}$ $\Big)$

x) Given expression: $\Big($ $\frac{x}{a}$ $-$ $\frac{a}{x}$ $\Big)^{10}$

Here $n = 10$ is even. Therefore $\Big($ $\frac{10}{2}$ $+1\Big)$ i.e. $6^{th}$ term is the middle term.

$T_6 = T_{5+1} = ^{10} C_5 \cdot \Big($ $\frac{x}{a}$ $\Big)^{10-5} \cdot \Big($ $\frac{-a}{x}$ $\Big)^5$

$= - {^{10} C_5} \cdot \Big($ $\frac{x}{a}$ $\Big)^5 \cdot \Big($ $\frac{a}{x}$ $\Big)^5$

$= - {^{10} C_5}$

$= -252$

$\\$

Question 16: Find the term independent of $x$ in the expansion of the following expressions:

i) $\Big($ $\frac{3}{2}$ $x^2 -$ $\frac{1}{3x}$ $\Big)^{9}$ ii) $\Big( 2x +$ $\frac{1}{3x^2}$ $\Big)^{9}$ iii) $\Big( 2x^2 -$ $\frac{3}{x^3}$ $\Big)^{25}$ iv) $\Big( 3x -$ $\frac{2}{x^2}$ $\Big)^{15}$

v) $\Big($ $\sqrt{\frac{x}{3}}$ $+$ $\frac{\sqrt{3}}{2x^2}$ $\Big)^{10}$ vi) $\Big( x -$ $\frac{1}{x^2}$ $\Big)^{3n}$ vii) $\Big($ $\frac{1}{2}$ $x^{\frac{1}{3}} + x^{-\frac{1}{5}} \Big)^{15}$

viii) $(1 + x + 2x^3) \Big($ $\frac{3}{2}$ $x^2 -$ $\frac{1}{3x}$ $\Big)^{9}$ ix) $\Big( \sqrt[3]{x} +$ $\frac{1}{2}$ $\sqrt[3]{x} \Big)^{18} , x > 2$ x) $\Big($ $\frac{3}{2}$ $x^2 -$ $\frac{1}{3x}$ $\Big)^{6}$

Answer:

i) Given expression: $\Big($ $\frac{3}{2}$ $x^2 -$ $\frac{1}{3x}$ $\Big)^{9}$

Let $T_{r+1}$ term be independent of $x$

Therefore $T_{r+1} = {^9C_{r}} \cdot \Big($ $\frac{3}{2}$ $x^2 \Big)^{9-r} \cdot \Big($ $\frac{-1}{3x}$ $\Big)^{r}$

$= (-1)^r \cdot {^9C_{r}} \cdot \Big($ $\frac{3}{2}$ $\Big)^{9-r} \cdot \Big($ $\frac{x^{18-2r}}{3^r x^r}$ $\Big)$

If $T_{r+1}$ is to be independent of $x$ , then

$18-2r-r=0 \Rightarrow r = 6$ Therefore the $7^{th}$ term.

Therefore the required term $T_{6+1} = (-1)^6 \cdot {^9C_{6}} \cdot \Big($ $\frac{3}{2}$ $\Big)^{9-6} \cdot$ $\frac{1}{3^6}$ $= {^9C_{6}} \cdot \Big($ $\frac{3^3}{2^3 3^6}$ $\Big) =$ $\frac{7}{18}$

$\\$

ii) Given expression: $\Big( 2x +$ $\frac{1}{3x^2}$ $\Big)^{9}$

Let $T_{r+1}$ term be independent of $x$

Therefore $T_{r+1} = {^9C_{r}} \cdot (2x)^{9-r} \cdot \Big($ $\frac{1}{3x^2}$ $\Big)^{r}$

$= {^9C_{r}} \cdot 2^{9-r} \cdot \Big($ $\frac{x^{9-r}}{3^r x^{2r}}$ $\Big)$

If $T_{r+1}$ is to be independent of $x$ , then

$9-r-2r=0 \Rightarrow r = 3$ Therefore the $4^{th}$ term.

Therefore the required term $T_{3+1} = ^9 C_3 \cdot \Big($ $\frac{2^6}{3^3}$ $\Big) =$ $\frac{64}{27}$ $\cdot ^9 C_3 =$ $\frac{1792}{9}$

$\\$

iii) Given expression: $\Big( 2x^2 -$ $\frac{3}{x^3}$ $\Big)^{25}$

Let $T_{r+1}$ term be independent of $x$

Therefore $T_{r+1} = {^{25}C_{r}} \cdot (2x^2)^{25-r} \cdot \Big($ $\frac{-3}{x^3}$ $\Big)^{r}$

$= (-3)^r \cdot {^{25}C_{r}} \cdot 2^{25-r} \cdot \Big($ $\frac{x^{50-2r}}{x^{3r}}$ $\Big)$

If $T_{r+1}$ is to be independent of $x$ , then

$50-2r-3r=0 \Rightarrow r = 10$ Therefore the ${11}^{th}$ term.

Therefore the required term $T_{10+1} =(-3)^{10} \cdot ^{25}C_{10} \cdot 2^{25-10} = 3^{10} \cdot 2^{15} \cdot ^{25}C_{10}$

$\\$

iv) Given expression: $\Big( 3x -$ $\frac{2}{x^2}$ $\Big)^{15}$

Let $T_{r+1}$ term be independent of $x$

Therefore $T_{r+1} = {^{15}C_{r}} \cdot (3x)^{15-r} \cdot \Big($ $\frac{-2}{x^2}$ $\Big)^{r}$

$= (-1)^r \cdot {^{15}C_{r}} \cdot 3^{15-r} \cdot \Big($ $\frac{x^{15-r}2^r}{x^{2r}}$ $\Big)$

If $T_{r+1}$ is to be independent of $x$ , then

$15-r-2r=0 \Rightarrow r = 5$ Therefore the ${6}^{th}$ term.

Therefore the required term $T_{5+1} = ((-1)^5 \cdot {^{15}C_{5}} \cdot 3^{15-5} \cdot 2^5 = - ^{15} C_5 \cdot 3^{10} \cdot 2^5 = - 3003 \cdot 3^{10} \cdot 2^5$

$\\$

v) Given expression: $\Big($ $\sqrt{\frac{x}{3}}$ $+$ $\frac{\sqrt{3}}{2x^2}$ $\Big)^{10}$

Let $T_{r+1}$ term be independent of $x$

Therefore $T_{r+1}= ^{10} C_{r} \cdot$ $( \sqrt{\frac{x}{3} })^{10-r} \cdot ( \frac{ \sqrt{3} }{2x^2} )^r$

$= {^{10}C_{r}} \cdot \Big($ $\frac{x^{\frac{10-r}{2} }}{3^{\frac{10-r}{2}} } \cdot \frac{ 3^{ \frac{r}{2} } }{2^r x^{2r}}$ $\Big)$

If $T_{r+1}$ is to be independent of $x$ , then

$\frac{10-r}{2}$ $-2r =0 \Rightarrow 10-r-4r = 0 \Rightarrow r = 2$ Therefore the ${3}^{rd}$ term.

Therefore the required term $T_{3} = {^{10}C_{2}} \Big($ $\frac{3^{ \frac{2}{2} }}{2^2 \cdot 3^{\frac{10-2}{2} }}$ $\Big) = {^{10} C_{2}} \Big($ $\frac{3}{4 \cdot 3^4}$ $\Big) =$ $\frac{5}{12}$

$\\$

vi) Given expression: $\Big( x -$ $\frac{1}{x^2}$ $\Big)^{3n}$

Let $T_{r+1}$ term be independent of $x$

$T_{r+1} = ^{3n} C_r \cdot (x)^{3n} \cdot \Big($ $\frac{-1}{x^2}$ $\Big)^r$

$= (-1)^r {^{3n} C_r} \cdot$ $\frac{x^{3n-r}}{x^{2r}}$

If $T_{r+1}$ is to be independent of $x$ , then,

$3n-r-2r \Rightarrow r = n$ i.e $(n+1)^{th}$ term

$T_{n+1} = ( -1)^n \cdot {^{3n} C_n}$

$\\$

vii) $\Big($ $\frac{1}{2}$ $x^{\frac{1}{3}} + x^{-\frac{1}{5}} \Big)^{15}$

Let $T_{r+1}$ term be independent of $x$

$T_{r+1} = ^8 C_r \Big($ $\frac{1}{2}$ $x^{\frac{1}{3}}$ $\Big)^{8-r} \cdot ( x^{\frac{-1}{5}})^r$

$= ^8 C_r \Big($ $\frac{x^{\frac{8-r}{3}}}{2^{8-r}}$ $\Big) \cdot x^{\frac{-r}{5}}$

If $T_{r+1}$ is to be independent of $x$ , then,

$\frac{8-r}{3}$ $+$ $\frac{-r}{5}$ $= 0 \Rightarrow 40-5r-3r = 0 \Rightarrow r = 5$ Therefore the ${6}^{th}$ term.

$T_{5+1} = ^8 C_5 \cdot$ $\frac{1}{2^{8-5}}$ $= ^8 C_5 \cdot$ $\frac{1}{8}$ $= 7$

$\\$

viii) Given expression: $(1 + x + 2x^3) \Big($ $\frac{3}{2}$ $x^2 -$ $\frac{1}{3x}$ $\Big)^{9}$

$= (1+x+2x^3) \Big[ ^9 C_0 \Big( \frac{3}{2} x^2 \Big)^{9-0} \Big( \frac{-1}{3x} \Big)^0 + ^9 C_1 \Big( \frac{3}{2} x^2 \Big)^{9-1} \Big( \frac{-1}{3x} \Big)^1 + \ldots + ^9 C_9 \Big( \frac{3}{2}x^2 \Big)^{9-9} \Big( \frac{-1}{3x} \Big)^9\Big]$

Now $T_{r+1} = ^9 C_r \cdot \Big($ $\frac{3}{2}$ $x^2 \Big)^{9-r} \cdot \Big($ $\frac{-1}{3x}$ $\Big)^r$

$= (-1)^r \cdot {^9 C_r} \cdot \Big($ $\frac{3}{2}$ $\Big)^{9-r} \cdot x^{18-2r} \cdot \Big($ $\frac{1}{3^r x^r}$ $\Big)$

$= (-1)^r \cdot {^9 C_r} \cdot \Big($ $\frac{(\frac{3}{2})^{9-r}}{3^r}$ $\Big) \cdot x^{18-2r}$

Let $T_{r+1}$ term be independent of $x$

$18-3r = 0 \Rightarrow r = 6$ . Therefore it is the $7^{th}$ term.

Coefficient of the $7^{th}$ term $= (-1)^6 \cdot {^9 C_6} \Big($ $\frac{(\frac{3}{2})^{9-6}}{3^6}$ $\Big) =$ $\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}$ $\times$ $\frac{3^3}{3^6 \cdot 2^3}$ $=$ $\frac{7}{18}$

Let $T_{r+1}$ term be independent of $x^{-1}$

$18-3r = -1 \Rightarrow r =$ $\frac{17}{3}$ . This is not possible. Hence there is no term with $x^{-1}$ in the expansion.

If $T_{r+1}$ is to be independent of $x^{-3}$ , then,

$18-3r = -3 \Rightarrow r = 7$ . Therefore it is the $8^{th}$ term.

Coefficient of the $8^{th}$ term

$= (-1)^7 \cdot {^9 C_7} \Big($ $\frac{(\frac{3}{2})^{9-7}}{3^7}$ $\Big) = (-1)$ $\frac{9 \cdot 8 }{ 2 \cdot 1}$ $\times$ $\frac{3^2}{3^7 \cdot 2^2}$ $= -$ $\frac{2}{27}$

Therefore the coefficient of the term independent of $x =$ $\frac{7}{18}$ $-$ $\frac{2}{27}$ $=$ $\frac{17}{54}$

$\\$

ix) Given expression: $\Big( \sqrt[3]{x} +$ $\frac{1}{2}$ $\sqrt[3]{x} \Big)^{18} , x > 2$

Let $T_{r+1}$ term be independent of $x$

$T_{r+1}= ^{18} C_r ( x^{\frac{1}{3}} )^{18-r} \Big($ $\frac{1}{2x^{\frac{1}{3}}}$ $\Big)^r$

$= ^{18} C_r \Big($ $\frac{x^{\frac{18-r}{3}}}{2^r - x^{\frac{r}{3}}}$ $\Big)$

If $T_{r+1}$ is to be independent of $x$ , then,

$\frac{18-r}{3}$ $-$ $\frac{r}{3}$ $= 0 \Rightarrow r = 9$ Therefore it is the ${10}^{th}$ term.

$T_{9+1} = ^{18} C_{9} \cdot$ $\frac{1}{2^9}$ $=$ $\frac{ 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{1}{2^9}$ $=$ $\frac{12155}{128}$

$\\$

x) Given expression: $\Big($ $\frac{3}{2}$ $x^2 -$ $\frac{1}{3x}$ $\Big)^{6}$

Let $T_{r+1}$ term be independent of $x$

$T_{r+1} = ^6C_r \Big($ $\frac{3}{2}$ $x^2 \Big)^{6-r} \cdot \Big($ $\frac{-1}{3x}$ $\Big)^r$

$= (-1)^r \cdot {^6C_r} \cdot \Big($ $\frac{3}{2}$ $\Big)^{6-r} \cdot \Big($ $\frac{x^{12-2r}}{3^r x^r}$ $\Big)$

If $T_{r+1}$ is to be independent of $x$ , then,

$12 - 2r - r = 0 \Rightarrow r = 4$ . Therefore it is the ${5}^{th}$ term.

$T_{4+1} = (-1)^4 \cdot {^6 C_4} \cdot \Big($ $\frac{3}{2}$ $\Big)^2 \cdot$ $\frac{1}{3^4}$ $=$ $\frac{6 \cdot 5}{2 \cdot 1}$ $\times$ $\frac{3^2}{2^2}$ $\times$ $\frac{1}{3^4}$ $=$ $\frac{5}{12}$

$\\$

Question 17: If the coefficients of $( 2r + 4)^{th}$ and $(r-2)^{th}$ terms in the expansion of $( 1 + x)^{18}$ are equal, find $r$ .

Answer:

$[ \text{Note: When } ^n C_x = ^n C_y \Rightarrow x+y = n ]$

Given expression: $( 1 + x)^{18}$

$T_{2r+4} = ^{18} C_{2r+3} \cdot (x)^{2r+3}$

$T_{r-2} = ^{18} C_{r-3} \cdot (x)^{r-3}$

Given $^{18} C_{2r+3} = ^{18} C_{r-3}$

$\Rightarrow 2r+3 + r-3 = 18 \Rightarrow 3r = 18 \Rightarrow r = 6$

$\\$

Question 18: If the coefficient of $(2r+1)^{th}$ and $(r + 2)^{th}$ term in the expansion of $( 1 + x)^{43}$ are equal, find $r$ .

Answer:

$[ \text{Note: When} ^n C_x = ^n C_y \Rightarrow x+y = n ]$

Given expression: $( 1 + x)^{43}$

$T_{2r+1} = ^{43} C_{2r} \cdot (x)^{2r}$

$T_{r+2} = ^{43} C_{r+1} \cdot (x)^{r+1}$

Given $^{43} C_{2r} = ^{43} C_{r+1}$

$\Rightarrow 2r+r+1 = 43 \Rightarrow 3r = 42 \Rightarrow r = 14$

$\\$

Question 19: Prove that the coefficient of $(r+1)^{th}$ term in the expansion of $( 1+x)^{n+1}$ is equal to the sum of the coefficients of the $r^{th}$ and $(r+1)^{th}$ terms in the expansion of $( 1+x)^n$ .

Answer:

For expression: $( 1+x)^{n+1}$

$T_{r+1} = ^{n+1} C_r (x)^r$

For expression: $(1+x)^n$

$T_{r+1} = ^n C_r ( x)^r$

$T_r = ^n C_{r-1} (x)^{r-1}$

Sum of the coefficients $= ^n C_r + ^n C_{r-1} = ^{n+1} C_r$

Hence proven.

$\\$

Question 20: Prove that the term independent of $x$ in the expansion of

$\Big( x+$ $\frac{1}{x}$ $\Big)^{2n}$ is $\frac{1\cdot 3\cdot 5 \ldots (2n-1)}{n!}$ $\cdot 2^n$ .

Answer:

Given expression: $\Big( x+$ $\frac{1}{x}$ $\Big)^{2n}$

$T_{r+1} = ^{2n} C_r (x)^{2n-r} \Big($ $\frac{1}{x}$ $\Big)^r$

$= ^{2n} C_r x^{2n-2r}$

If the term is independent of $x$ , then

$2n-2r = 0 \Rightarrow r = n$

$\therefore T_{r+1} = ^{2n} C_n =$ $\frac{(2n)!}{ n! n!}$

$=$ $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots (2n-2) \cdot (2n-1) \cdot (2n)}{n! n!}$

$=$ $\frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)][ 2 \cdot 4 \cdot 6 \cdot 8 \ldots (2n-2) \cdot (2n)}{n! n!}$

$=$ $\frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)] 2^n [ 1 \cdot 2 \cdot 3 \cdot 4 \ldots (n-1) \cdot (n)]}{n! n!}$

$=$ $\frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)] 2^n}{n!}$

Therefore the term independent of $x$ is $\frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)] 2^n}{n!}$

$\\$

Question 21: The coefficient of the $5^{th}, 6^{th}$ and $7^{th}$ terms in the expansion of $(1+x)^n$ are in A.P. , find $n$ .

Answer:

Given expression: $(1+x)^n$

Coefficient of $5^{th} \text{term} = ^n C_{5-1} = ^n C_4$

Coefficient of $6^{th} \text{term} = ^n C_{6-1} = ^n C_5$

Coefficient of $7^{th} \text{term} = ^n C_{7-1} = ^n C_6$

Since they are in AP

$2 \cdot ^n C_5 = ^n C_4 + ^n C_6$

$\Rightarrow 2 \cdot$ $\frac{n!}{5! (n-5)!}$ $=$ $\frac{n!}{4! (n-4)!}$ $+$ $\frac{n!}{6! (n-6)!}$

$\Rightarrow$ $\frac{2}{5 \cdot 4! (n-5) (n-6)!}$ $=$ $\frac{1}{4! (n-4)(n-5) (n-6)!}$ $+$ $\frac{1}{6 \cdot 5 \cdot 4! (n-6)!}$

$\Rightarrow$ $\frac{1}{(n-1)} \Big[$ $\frac{2}{5}$ $-$ $\frac{1}{(n-4)}$ $\Big] =$ $\frac{1}{30}$

$\Rightarrow$ $\frac{1}{(n-1)}$ $\Big[$ $\frac{2n - 8 - 5}{5 (n-4)}$ $\Big] =$ $\frac{1}{30}$

$\Rightarrow 6(2n-13) = (n-4)(n-5)$

$\Rightarrow 12n - 78 = n^2 - 9n +20$

$\Rightarrow n^2 - 21 n + 108 = 0$

$\Rightarrow (n-7)(n-14) = 0$

$\Rightarrow n = 7 \ or \ n = 14$

$\\$

Question 22: If the coefficient of $2^{nd}, 3^{rd}$ and $4^{th}$ terms in the expansion of $(1+x)^{2n}$ are in A.P., show that $2n^2 - 9n + 7 = 0$ .

Answer:

Given expression: $(1+x)^{2n}$

$T_{1+1} = ^{2n} C_1 \cdot x^1$

$T_{2+1} = ^{2n} C_2 \cdot x^2$

$T_{3+1} = ^{2n} C_3 \cdot x^3$

Therefore $2 ( ^{2n} C_2 ) = ^{2n} C_1 + ^{2n} C_3$

$\Rightarrow 2 \cdot$ $\frac{(2n)!}{2! (2n-2)!}$ $=$ $\frac{(2n)!}{1! (2n-1)!}$ $+$ $\frac{(2n)!}{3! (2n-3)!}$

$\Rightarrow$ $\frac{2}{2 (2n-2)(2n-3)!}$ $=$ $\frac{1}{(2n-1)(2n-2)(2n-3)!}$ $+$ $\frac{1}{3 \cdot 2 \cdot 1 ( 2n-3)!}$

$\Rightarrow$ $\frac{1}{2n-2}$ $=$ $\frac{1}{(2n-1)(2n-2)}$ $+$ $\frac{1}{6}$

$\Rightarrow$ $\frac{1}{(2n-2)}$ $\Big[ 1 -$ $\frac{1}{2n-1}$ $\Big] =$ $\frac{1}{6}$

$\Rightarrow$ $\frac{1}{(2n-2)}$ $\Big[$ $\frac{2n-1-1}{2n-1}$ $\Big] =$ $\frac{1}{6}$

$\Rightarrow 6(2n-2) = ( 2n-2)(2n-1)$

$\Rightarrow 12n - 12 = 4n^2 - 4n - 2n +2$

$\Rightarrow 4n^2 - 18n +14 = 0$

$\Rightarrow 2n^2 -9n+7 = 0$ Hence proved.

$\\$

Question 23: If the coefficient of $2^{nd}, 3^{rd}$ and $4^{th}$ terms in the expansion of $(1+x)^{n}$ are in A.P., then find the value on $n$ .

Answer:

Given expression: $(1+x)^{n}$

$T_{1+1} = ^{n} C_1 \cdot x^1$

$T_{2+1} = ^{n} C_2 \cdot x^2$

$T_{3+1} = ^{n} C_3 \cdot x^3$

Therefore $2 ( ^{n} C_2 ) = ^{n} C_1 + ^{n} C_3$

$\Rightarrow 2 \cdot$ $\frac{n!}{2! (n-2)!}$ $=$ $\frac{n!}{1! (n-1)!}$ $+$ $\frac{n!}{3! (n-3)!}$

$\Rightarrow$ $\frac{1}{ (n-2)(n-3)!}$ $=$ $\frac{1}{1! (n-1)(n-2)(n-3)!}$ $+$ $\frac{1}{3 \cdot 2 \cdot 1 ( n-3)!}$

$\Rightarrow$ $\frac{1}{n-2}$ $=$ $\frac{1}{(n-1)(n-2)}$ $+$ $\frac{1}{6}$

$\Rightarrow$ $\frac{1}{(n-2)}$ $\Big[ 1 -$ $\frac{1}{n-1}$ $\Big] =$ $\frac{1}{6}$

$\Rightarrow$ $\frac{1}{(n-2)}$ $\Big[$ $\frac{n-1-1}{n-1}$ $\Big] =$ $\frac{1}{6}$

$\Rightarrow 6(n-2) = ( n-2)(n-1)$

$\Rightarrow (n-2)[ 6 - (n-1)]= 0$

$\Rightarrow (n-2)(7-n) = 0$

$\Rightarrow n = 2 \text{ or } n = 7$

$n = 2$ is not possible as then $2 > 3$ in the $4^{th}$ term.

Hence $n = 7$ .

$\\$

Question 24: If in the expansion of $(1+x)^n$ , the coefficients of $p^{th}$ and $q^{th}$ terms are equal, prove the $p+q = n + 2$ , where $p \neq q$ .

Answer:

Given expression: $(1+x)^n$

$T_p = T _{(p-1)+1} = ^n C_{p-1} x^{p-1}$

$T_q = T _{(q-1)+1} = ^n C_{q-1} x^{q-1}$

$\therefore ^n C_{p-1} = ^n C_{q-1}$

$\Rightarrow (p-1) + (q-1) = n$

$\Rightarrow p+q = n + 2$ . Hence proved.

$\\$

Question 25: Find $a$ , if the coefficients of $x^2$ and $x^3$ in the expansion of $(3 + ax)^9$ are equal.

Answer:

Given expression:

$(3 + ax)^9= ^9 C_0 (3)^9 (ax)^0 + ^9 C_1 (3)^8 (ax)^1 +^9 C_2 (3)^7 (ax)^2 +^9 C_3 (3)^6 (ax)^3 + \ldots$

Therefore coefficient of $x^2 = ^9 C_2 (3)^7 a^2$

and coefficient of $x^3 = ^9 C_3 (3)^6 a^3$

$\therefore ^9 C_2 (3)^7 a^2 = ^9 C_3 (3)^6 a^3$

$\frac{9!}{2! 7!}$ $\cdot 3^7 =$ $\frac{9!}{3! 6!}$ $\cdot 3^6 a$

$a =$ $\frac{3\times 3}{7}$ $=$ $\frac{9}{7}$

$\\$

Question 26: Find the coefficient of $a^4$ in the product $(1 + 2a)^4 ( 2 - a)^5$ using binomial theorem.

Answer:

Given expression:

$(1 + 2a)^4 ( 2 - a)^5$

$= \Big[ ^4 C_0 (2a)^0+^4 C_1 (2a)^1+^4 C_2 (2a)^2+^4 C_3 (2a)^3+^4 C_4 (2a)^4 \Big] \times \Big[ ^5 C_0 (2)^5 (-a)^0 \\ \\ + ^5 C_1 (2)^4 (-a)^1 + ^5 C_2 (2)^3 (-a)^2 + ^5 C_3 (2)^2 (-a)^3 + ^5 C_4 (2)^1 (-a)^4 + ^5 C_5 (2)^0 (-a)^5 \Big]$

$= [ 1+8a + 24a^2 + 32a^3 + 16a^4 ] \times [ 32-80a+80a^2 - 40a^3 + 10a^4 - a^5 ]$

$\therefore$ coefficient of $a^4 = 10 + (8)(-40) + (24)(80) + (32)(-80) + (16)(32)$

$= 10 - 320 + 1920 - 2560 + 512$

$= - 438$

$\\$

Question 27: In the expansion of $(1+x)^n$ the binomial coefficients of three consecutive terms are respectively $220, 495$ and $792$ , find the value of $n$ .

Answer:

Let the consecutive terms be $T_{r-1}, \ \ T_r, \ \ T_{r+1}$

The binomial coefficient for these terms will be $^n C_{r-2}, \ \ ^n C_{r-1}, \ \ ^n C_{r}$ , respectively.

It is given, $^n C_{r-2} = 220, ^n C_{r-1} = 495$ and $^n C_{r} = 792$

$\therefore$ $\frac{^n C_{r-2}}{^n C_{r-1}}$ $=$ $\frac{220}{495}$

$\Rightarrow$ $\frac{r-1}{n-r+2}$ $=$ $\frac{4}{9}$

$\Rightarrow 9r - 9 = 4n - 4r + 8$

$\Rightarrow 4n + 17 = 13r$ … … … … … i)

Similarly, $\frac{^n C_{r}}{^n C_{r-1}}$ $=$ $\frac{792}{495}$

$\Rightarrow$ $\frac{n-r+1}{r}$ $=$ $\frac{6}{5}$

$\Rightarrow 5n - 5r + 7 = 8r$

$\Rightarrow 5n + 5 = 13 r$ … … … … … ii)

From i) and ii) we get

$4n + 17 = 5n + 5$

$\Rightarrow n = 12$

$\\$

Question 28: In the expansion of $(1+x)^n$ the coefficients of three consecutive terms are respectively $56, 76$ and $56$ , then find $n$ and the position of the terms of these coefficients.

Answer:

Let the consecutive terms be $T_{r}, \ \ T_{r+1}, \ \ T_{r+2}$

The binomial coefficient for these terms will be $^n C_{r-1}, \ \ ^n C_{r}, \ \ ^n C_{r+1}$ , respectively.

We have $^n C_{r-1} = ^n C_{r+1} = 56$

$\Rightarrow r - 1 + r + 1 = n$

$\Rightarrow 2r = n$

$\Rightarrow r = \frac{n}{2}$

Now, $^n C_{\frac{n}{2}} = 70$ and $^n C_{\frac{n}{2}-1} = 56$

Dividing $\frac{^n C_{\frac{n}{2}-1}}{^n C_{\frac{n}{2}}}$ $=$ $\frac{56}{70}$

$\Rightarrow$ $\frac{n!}{\big( \frac{n}{2} -1\big) \big( \frac{n}{2} + 1\big) } \times \frac{\big( \frac{n}{2} \big) \big( \frac{n}{2} \big) }{n!}$ $=$ $\frac{8}{10}$

$\Rightarrow$ $\frac{ \big( \frac{n}{2} \big) \big( \frac{n}{2} -1\big) \big( \frac{n}{2} \big) }{ \big(\frac{n}{2}\big) \big( \frac{n}{2} -1\big) \big( \frac{n}{2} + 1 \big)}$ $=$ $\frac{8}{10}$

$\Rightarrow$ $\frac{\frac{n}{2} }{\frac{n}{2}+1}$ $=$ $\frac{8}{10}$

$\Rightarrow 5n = 4n + 8$

$\Rightarrow n = 8$

$\therefore r =$ $\frac{n}{2}$ $=$ $\frac{8}{2}$ $= 4$

Therefore the required terms are $4^{th}, \ 5^{th}$ and $6^{th}$ .

$\\$

Question 29: If $3^{rd}, 4^{th}, 5^{th}$ and $6^{th}$ terms in the expansion $( x+ \alpha)^n$ be respectively $a, b, c$ and $d$ , prove that $\frac{b^2-ac}{c^2-bd} = \frac{5a}{3c}$ .

Answer:

Given expression: $( x+ \alpha)^n$

$T_3 = ^n C_2 \ (x)^{n-2} \ (\alpha)^2$ $T_4 = ^n C_3 \ (x)^{n-3} \ (\alpha)^3$

$T_5 = ^n C_4 \ (x)^{n-4} \ (\alpha)^4$ $T_6 = ^n C_5 \ (x)^{n-5} \ (\alpha)^5$

Given:

$^n C_2 \ (x)^{n-2} \ (\alpha)^2 = a$ $^n C_3 \ (x)^{n-3} (\alpha)^3 = b$

$^n C_4 \ (x)^{n-4} \ (\alpha)^4 = c$ $^n C_5 \ (x)^{n-5} (\alpha)^5 = d$

We have to prove:

$\frac{b^2-ac}{c^2-bd} = \frac{5a}{3c}$

$\Rightarrow$ $\frac{b^2 - ac}{a} = \frac{5}{3} \Big[ \frac{c^2- bd}{c} \Big]$

$\Rightarrow$ $\frac{1}{b} \Big[ \frac{b^2 - ac}{a} \Big] = \frac{5}{3} \Big[ \frac{c^2- bd}{c} \Big]$

$\Rightarrow$ $\frac{b}{a} - \frac{c}{b} = \frac{5}{3} \Big[ \frac{c}{d} - \frac{d}{c} \Big]$ … … … … … i)

Substituting the values in i) we get

$\frac{^n C_3 (x)^{n-3} (\alpha)^3}{^n C_2 (x)^{n-2} (\alpha)^2} - \frac{^n C_4 (x)^{n-4} (\alpha)^4}{^n C_3 (x)^{n-3} (\alpha)^3} = \frac{5}{3} \Big[ \frac{^n C_4 (x)^{n-4} (\alpha)^4}{^n C_3 (x)^{n-3} (\alpha)^3} - \frac{^n C_5 (x)^{n-5} (\alpha)^5}{^n C_4 (x)^{n-4} (\alpha)^4} \Big]$

$\Rightarrow$ $\Big[ \frac{^n C_3}{^n C_2} - \frac{^n C_4}{^n C_3} \Big] \frac{ \alpha }{x}= \frac{5 \alpha}{3 x} \Big[ \frac{^n C_4}{^n C_3} - \frac{^n C_5}{^n C_4} \Big]$

We know $\frac{^n C_r}{^n C_{r-1}} = \frac{n-r+1}{r}$

$\Rightarrow$ $\Big[ \frac{n-2}{3} - \frac{n-3}{4} \Big] = \frac{4}{3} \Big[ \frac{n-3}{4} - \frac{n-4}{5} \Big]$

$\Rightarrow$ $\frac{4n-8-3n+9}{3 \times 4} = \frac{5n - 15 - 4n + 16}{3 \times 4}$

$\Rightarrow$ $\frac{n+1}{12} = \frac{n+1}{12}$

Therefore LHS $=$ RHS. Hence proved.

$\\$

Question 30: If $a, b, c$ and $d$ in any binomial expansion be the $6^{th}, 7^{th}, 8^{th}$ and $9^{th}$ terms respectively, then prove that $\frac{b^2-ac}{c-bd} = \frac{4a}{3c}$ .

Answer:

Let the expression be $(x + \alpha)^n$

Given: $T_6 = a, \ \ T_7 = b, \ \ T_8 = c, \ \ T_9 = d$

We have to prove:

$\frac{b^2-ac}{c^2-bd} = \frac{4a}{3c}$

$\Rightarrow$ $\frac{b^2 - ac}{a} = \frac{4}{3} \Big[ \frac{c^2- bd}{c} \Big]$

$\Rightarrow$ $\frac{b}{a} - \frac{c}{b} = \frac{4}{3} \Big[ \frac{c}{d} - \frac{d}{c} \Big]$ … … … … … i)

We know:

$a = ^n C_5 \ x^{n-5} \ {\alpha}^5$ $b = ^n C_6 \ x^{n-6} \ {\alpha}^6$

$c = ^n C_7 \ x^{n-7} \ {\alpha}^7$ $d = ^n C_8 \ x^{n-8} \ {\alpha}^8$

Substituting in i) we get

$\frac{^n C_6 \ x^{n-6} \ {\alpha}^6}{^n C_5 \ x^{n-5} \ {\alpha}^5} - \frac{^n C_7 \ x^{n-7} \ {\alpha}^7}{^n C_6 \ x^{n-6} \ {\alpha}^6} = \frac{4}{3} \Big[ \frac{^n C_7 \ x^{n-7} \ {\alpha}^7}{^n C_6 \ x^{n-6} \ {\alpha}^6} - \frac{^n C_8 \ x^{n-8} \ {\alpha}^8}{^n C_7 \ x^{n-7} \ {\alpha}^7} \Big]$

$\Rightarrow$ $\Big[ \frac{^n C_6}{^n C_5} - \frac{^n C_7}{^n C_6} \Big] \frac{ \alpha }{x}= \frac{4 \alpha}{3 x} \Big[ \frac{^n C_7}{^n C_6} - \frac{^n C_8}{^n C_7} \Big]$

We know $\frac{^n C_r}{^n C_{r-1}} = \frac{n-r+1}{r}$

$\Rightarrow$ $\Big[ \frac{n-5}{6} - \frac{n-6}{7} \Big] = \frac{4}{3} \Big[ \frac{n-6}{7} - \frac{n-7}{8} \Big]$

$\Rightarrow$ $\frac{7n-35-6n+36}{6 \times 7} = \frac{8n-48-7n+49}{3 \times 7 \times 2}$

$\Rightarrow$ $\frac{n+1}{42} = \frac{n+1}{42}$

Therefore LHS $=$ RHS. Hence proved.

$\\$

Question 31: If the coefficient of three consecutive terms in the expansion $( 1+x)^n$ be $76, 95$ and $76$ , find $n$ .

Answer:

Let the three consecutive terms be $T_r, T_{r+1}, T_{r+2}$

Coefficient of $T_r = ^n C_{r-1} = 76$

Coefficient of $T_{r+1} = ^n C_{r} = 95$

Coefficient of $T_{r+2} = ^n C_{r+1} = 76$

Now, $\frac{^n C_{r+1}}{^n C_{r}} = \frac{76}{95}$

$\Rightarrow$ $\frac{n - ( r+1) + 1}{r+1}= \frac{76}{95}$

$\Rightarrow$ $\frac{n-r}{r+1}= \frac{4}{5}$

$\Rightarrow 5n - 5r = 4r + 4$

$\Rightarrow 5n - 9r = 4$ … … … … … i)

Similarly, $\frac{^n C_{r}}{^n C_{r-1}} = \frac{95}{76}$

$\Rightarrow$ $\frac{n-r+1}{r}= \frac{5}{4}$

$\Rightarrow 4n - 4r + 4= 5r$

$\Rightarrow 4n-9r=-4$ … … … … … ii)

Subtracting ii) from i) we get $n = 4+4 = 8$

$\\$

Question 32: If the $6^{th}, 7^{th}$ and $8^{th}$ in the expansion of $(x+a)^n$ are respectively $112, 7$ and $\frac{1}{4}$ , find $x, a, n$ .

Answer:

Given expression: $(x+a)^n$

Given $T_6 = 112, T_7 = 7$ and $T_8 =$ $\frac{1}{4}$

$T_6 = ^n C_5 \ x^{n-5} \ a^5 = 112$

$T_7 = ^n C_6 \ x^{n-6} \ a^6 = 7$

$T_8 = ^n C_7 \ x^{n-7} \ a^7 =$ $\frac{1}{4}$

Now, $\frac{T_7}{T_6} = \frac{^n C_6 \ x^{n-6} \ a^6}{^n C_5 \ x^{n-5} \ a^5} = \frac{7}{112}$

$\Rightarrow$ $\frac{^n C_6}{^n C_5} \times \frac{a}{x} = \frac{1}{16}$

$\Rightarrow$ $\frac{n-5}{6} \times \frac{a}{x} = \frac{1}{16}$

$\Rightarrow$ $\frac{a}{x} = \frac{3}{8n-40}$ … … … … … i)

Similarly, $\frac{T_8}{T_7} = \frac{^n C_7 \ x^{n-7} \ a^7}{^n C_6 \ x^{n-6} \ a^6} = \frac{\frac{1}{4}}{7}$

$\Rightarrow$ $\frac{^n C_7}{^n C_6} \times \frac{a}{x} = \frac{1}{28}$

$\Rightarrow$ $\frac{n-6}{7} \times \frac{a}{x} = \frac{1}{28}$

$\Rightarrow$ $\frac{a}{x} = \frac{1}{4n-24}$ … … … … … ii)

From i) and ii)

$\frac{3}{8n-40} = \frac{1}{4n-24}$

$\Rightarrow 12n - 72 = 8n -40$

$\Rightarrow 4n = 32$

$\Rightarrow n = 8$

Therefore $\frac{a}{x} = \frac{3}{24} = \frac{1}{8}$

Substituting we get,

$^8 C_5 \ x^{8-5} \ \Big($ $\frac{x}{8}$ $\Big)^5 = 112$

$\Rightarrow$ $\frac{56}{8}$ $x^8 = 112$

$\Rightarrow x^8 = 4^8$

$\Rightarrow x = 4$

Substituting in i) we get $a =$ $\frac{1}{2}$

$\\$

Question 33: If the $2^{nd}, 3^{rd}$ and $4^{th}$ term in the expansion of $(x+a)^n$ are $240, 720$ and $1080$ respectively, find $x, a, n$ .

Answer:

Given expression: $(x+a)^n$

Given $T_2 = 240, T_3 = 720$ and $T_4 = 1080$

$T_2 = ^n C_1 \ x^{n-1} \ a^1 = 240$

$T_3 = ^n C_2 \ x^{n-2} \ a^2 = 720$

$T_4 = ^n C_3 \ x^{n-3} \ a^3 = 1080$

Now, $\frac{T_3}{T_2} = \frac{^n C_2 \ x^{n-2} \ a^2}{^n C_1 \ x^{n-1} \ a^1} = \frac{720}{240}$

$\Rightarrow$ $\frac{n-1}{2x}$ $\cdot a = 3$

$\Rightarrow$ $\frac{a}{x} = \frac{6}{n-1}$ … … … … … i)

Similarly, $\frac{T_4}{T_3} = \frac{^n C_3 \ x^{n-3} \ a^3}{^n C_2 \ x^{n-2} \ a^2} = \frac{1080}{720}$

$\Rightarrow$ $\frac{n-2}{3x}$ $\cdot a = \frac{3}{2}$

$\Rightarrow$ $\frac{a}{x} = \frac{9}{2n-4}$ … … … … … ii)

From i) and ii) we get

$\frac{6}{n-1} = \frac{9}{2n-4}$

$\Rightarrow 12 n - 24 = 9n - 9$

$\Rightarrow 3n = 15$

$\Rightarrow n = 5$

Substituting in i) we get

$2a = 3x$ … … … … … iii)

Now $^n C_1 \ x^{5-1} \ \Big($ $\frac{3}{2}$ $x \Big) = 240$

$\Rightarrow 15x^5 = 480$

$\Rightarrow x^5 = 32 = 2^5$

$\Rightarrow x = 2$

From iii) $a =$ $\frac{3}{2}$ $\times 2 = 3$

$\\$

Question 34: Find $a, b$ and $n$ in the expansion $(a+b)^n$ , if the first three terms in the expansion are $729, 7290$ and $30375$ respectively.

Answer:

Given expression: $(a+b)^n$

Given: $T_1 = 729 \ , T_2 = 7290 \ , T_3 = 30375$

$T_1 = ^n C_0 \ a^n \ b^0 = 729 \Rightarrow a^n = 729 \Rightarrow a^n = 3^6$

Similarly, $T_2 = ^n C_1 \ a^{n-1} \ b^1 = 7290$

$T_3 = ^n C_2 \ a^{n-2} \ b^2 = 30375$

$\therefore \frac{T_3}{T_2} = \frac{^n C_2 \ a^{n-2} \ b^2}{^n C_1 \ a^{n-1} \ b^1} = \frac{30375}{7290}$

$\Rightarrow$ $\big( \frac{n-1}{2} \big) \big( \frac{b}{a} \big) = \frac{25}{3}$ … … … … … i)

Similarly, $\therefore \frac{T_2}{T_1} = \frac{^n C_1 \ a^{n-1} \ b^1}{^n C_0 \ a^{n} \ b^0} = \frac{7290}{729}$

$\Rightarrow$ $\big( \frac{nb}{a} \big) = 10$ … … … … … ii)

Dividing ii) by i) we get

$\frac{\frac{nb}{a}}{\frac{(n-1)b}{a}}$ $= 10 \times$ $\frac{3}{25}$

$\Rightarrow$ $\frac{n}{n-1} = \frac{6}{5}$

$\Rightarrow 5n = 6n - 6$

$\Rightarrow n = 6$

Since $a^6 = 3^6$

$\Rightarrow a = 3$

$\therefore$ $\frac{nb}{a}$ $= 10$

$\Rightarrow b = 10 \times$ $\frac{3}{6}$ $= 5$

$\\$

Question 35: If the term free from $x$ in the expansion $\Big( \sqrt{x} -$ $\frac{k}{x^2}$ $\Big)^{10}$ is $405$ , find the value of $k$ .

Answer:

Given expression: $\Big( \sqrt{x} -$ $\frac{k}{x^2}$ $\Big)^{10}$

Let $T_{r+1}$ term is independent of $x$

Therefore $T_{r+1} = ^{10} C_r \cdot (\sqrt{x})^{10-r} \cdot \Big($ $\frac{-k}{x^2}$ $\Big)^r$

$= (-1)^r \cdot {^{10} C_r} \cdot x^{\frac{10-r}{2}} \cdot$ $\frac{k^r}{x^{2r}}$

If $T_{r+1}$ is independent of $x$ , then

$\frac{10-r}{2} - 2r = 0 \Rightarrow 10-r-4r = 0 \Rightarrow r = 2$ . Therefore it is the $3^{rd}$ term.

$\therefore (-1)^2 \cdot {^{10} C_2} \cdot k^2 = 405$

$\Rightarrow$ $\frac{10 \cdot 9}{2 \cdot 1}$ $k^2 = 405$

$\Rightarrow k^2 =$ $\frac{405 \times 2}{10 \times 9}$ $= 9$

$\Rightarrow k = \pm 3$

$\\$

Question 36: Find the $6^{th}$ term in the expansion $\big( y^{\frac{1}{2}} + x^{\frac{1}{3}} \big)^n$ , if the binomial coefficient of the third term from the end is $45$ .

Answer:

Given expression: $\big( y^{\frac{1}{2}} + x^{\frac{1}{3}}\big)^n$

Third term from the end for $\big( y^{\frac{1}{2}} + x^{\frac{1}{3}} \big)^n$ is third term from the beginning for the expression $\big( x^{\frac{1}{3}} + y^{\frac{1}{2}} \big)^n$

$T_3 = ^n C_2 \cdot \big( x^{\frac{1}{3}} )^{n-2} \cdot ( y^{\frac{1}{2}} \big)^2$

Given $^n C_2 = 45$

$\Rightarrow$ $\frac{n(n-1)}{2}$ $= 45$

$\Rightarrow n(n-1) = 90$

$\Rightarrow n^2 - n - 90 = 0$

$\Rightarrow (n-10)(n+9) = 0$

$\Rightarrow n = 10$ or $n = -9$ (not possible as $n$ cannot ne negative)

$\therefore n = 10$

Therefore $\Rightarrow T_6 = ^{10} C_5 \cdot (y^{\frac{1}{2}})^{10-5} \cdot (x^{\frac{1}{3}})^5$

$\Rightarrow = ^{10} C_5 \cdot y^{\frac{5}{2}} \cdot x^{\frac{5}{3}}$

$\Rightarrow = 252 \cdot y^{\frac{5}{2}} \cdot x^{\frac{5}{3}}$

Therefore the $\Rightarrow 6^{th}$ term from the beginning is $252 \cdot y^{\frac{5}{2}} \cdot x^{\frac{5}{3}}$

$\\$

Question 37: If $p$ is a real number and if the middle term in the expansion of $\Big($ $\frac{p}{2}$ $+ 2 \Big)^{8}$ is $1120$ , find $p$ .

Answer:

Given expression: $\Big($ $\frac{p}{2}$ $+ 2 \Big)^{8}$

Middle term $= \Big($ $\frac{8}{2}$ $+1 \Big)$ i.e. $5^{th}$ term

$T_5 = ^8 C_4 \Big($ $\frac{p}{2}$ $\Big)^{8-4} (2)^4 = 1120$

$\therefore$ $\frac{ 8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}$ $\times$ $\frac{p^4}{2^4}$ $\times 2^4 = 1120$

$\Rightarrow p^4 =$ $\frac{1120}{7 \times 2 \times 5 }$ $= 16 = 2^4$

$\therefore p= \pm 2$

$\\$

Question 38: Find $n$ in the binomial $\Big( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \Big)^n$ , if the ratio of $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $\frac{1}{6}$ .

Answer:

Given expression: $\Big( \sqrt[3]{2} +$ $\frac{1}{\sqrt[3]{3}}$ $\Big)^n$

$7^{th}$ term from the end $= [(n+1)-7+1] = (n-5)^{th}$ term from the beginning.

$T_7 = ^n C_6 \cdot (\sqrt[3]{2})^{n-6} \cdot \Big($ $\frac{1}{\sqrt[3]{3}}$ $\Big)^6$

$= ^n C_6 \Big($ $\frac{2^{\frac{n-6}{3}}}{3^2}$ $\Big)$

$T_{n-5} = ^n C_{n-6} \Big(2^{\frac{1}{3}} \Big)^{n - ( n - 6)} \Big($ $\frac{1}{3^{\frac{1}{3}}}$ $\Big)^{n-6}$

$= ^n C_{n-6} \cdot 2^2 \cdot \Big($ $\frac{1}{3{\frac{n-6}{3}}}$ $\Big)$

Given $\frac{T_7}{T_{n-1} } = \frac{1}{6}$

$\Rightarrow$ $\frac{^n C_6 \Big( \frac{2^{\frac{n-6}{3} } }{3^2}\Big) }{^n C_{n-6} \Big( \frac{2^2}{3^{ \frac{n-6}{3} } } \Big) } = \frac{1}{6}$

$\Rightarrow$ $\frac{n!}{6!(n-6)!} \times \frac{(n-6)! 6!}{n!} \cdot 2^{{\frac{n-6}{3}-2} } \cdot 3^{\frac{n-6}{3}-2} =\frac{1}{6}$

$\Rightarrow 2^{\frac{n-12}{3}} \times 3^{ \frac{n-12}{3}} =$ $\frac{1}{6}$

$\Rightarrow 6^{\frac{n-12}{3}} = 6^{-1}$

$\Rightarrow$ $\frac{n-12}{3}$ $= -1$

$\Rightarrow n = 9$

$\\$

Question 39: If the $7^{th}$ term from the beginning and from the end in the binomial expansion $\Big( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \Big)^n$ are equal, find $n$ .

Answer:

Given expression: $\Big( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \Big)^n$

$7^{th}$ term from the end $= [(n+1)-7+1] = (n-5)^{th}$ term from the beginning.

$T_7 = ^n C_6 \cdot (\sqrt[3]{2})^{n-6} \cdot \Big($ $\frac{1}{\sqrt[3]{3}}$ $\Big)^6$

$= ^n C_6 \Big($ $\frac{2^{\frac{n-6}{3}}}{3^2}$ $\Big)$

$T_{n-5} = ^n C_{n-6} \Big(2^{\frac{1}{3}} \Big)^{n - ( n - 6)} \Big($ $\frac{1}{3^{\frac{1}{3}}}$ $\Big)^{n-6}$

$= ^n C_{n-6} \cdot 2^2 \cdot \Big($ $\frac{1}{3{\frac{n-6}{3}}}$ $\Big)$

Since, $T_7 = T_{n-5}$

$\Rightarrow ^n C_6 \Big($ $\frac{2^{\frac{n-6}{3}}}{3^2}$ $\Big) = ^n C_{n-6} \Big($ $\frac{2^2}{3^{\frac{n-6}{3}}}$ $\Big)$

$\Rightarrow 6^{\frac{n-6}{3}} = 6^2$

$\Rightarrow$ $\frac{n-6}{3}$ $= 2$

$\Rightarrow n = 12$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 11: Binomial Theorem – Exercise 18.2

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