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)

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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}}

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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}

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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}

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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}

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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

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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}  

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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}

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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}

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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} 

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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 

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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}

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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})!}  

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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

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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

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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

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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.

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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 .

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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} .

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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

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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

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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

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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}

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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}

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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}

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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

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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}

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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}

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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

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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}

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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}

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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}

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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

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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

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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.

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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!}

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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 

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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.

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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 .

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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.

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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}

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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

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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

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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} .

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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.

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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.

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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

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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}

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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

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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

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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

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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}}

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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

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

\Rightarrow n = 9

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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