Question 1: Using binomial theorem, write down the expansions of the following:

i) $(2x+3y)^5$        ii) $(2x-3y)^4$        iii) $\Big( x -$ $\frac{1}{x}$ $\Big)^6$        iv) $(1-3x)^7$

v) $\Big( ax -$ $\frac{b}{x}$ $\Big)^6$        vi) $\Big($ $\sqrt{\frac{x}{a}}$ $-$ $\sqrt{ \frac{a}{x}}$ $\Big)^6$        vii) $( \sqrt[3]{x} - \sqrt[3]{a} )^6$

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

i) $(2x+3y)^5$

$= ^5 C_0 (2x)^5 (3y)^0 + ^5 C_1 (2x)^4 (3y)^1 + \ldots + ^5 C_0 (2x)^0 (3y)^5$

$= 32x^5 + 5 \times 16 x^4 \times 3y + 10 \times 8x^3 \times 9y^2 + 10 \times 4x^2 \times 27y^3 + 5 \times 3x \times 81y^2 + 243y^5$

$= 32x^5 + 240x^4y + 720x^3y^2 + 1080 x^2 y^3 + 810 xy^4 + 243 y^5$

$\\$

ii) $(2x-3y)^4$

$= ^4 C_0 (2x)^4 (3y)^0 - ^4 C_1 (2x)^3 (3y)^1 + ^4 C_2 (2x)^2 (3y)^2 - ^4 C_3 (2x)^1 (3y)^3 +^4 C_4 (2x)^0 (3y)^4$

$= 16x^4 - 4 \times 8x^3 \times 3y + 6 \times 4x^2 \times 9y^2 - 4 \times 2x \times 27y^3 + 81y^4$

$= 16x^4 - 96 x^3 y + 216 x^2y^2 - 216 xy^3 + 81y^4$

$\\$

iii) $\Big( x -$ $\frac{1}{x}$ $\Big)^6$

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

$= x^6 - 6x^5 \times$ $\frac{1}{x}$ $+ 15x^4 \times$ $\frac{1}{x^2}$ $- 20x^3 \times$ $\frac{1}{x^3}$ $+ 15 x^2 \times$ $\frac{1}{x^4}$ $- 6x \times$ $\frac{1}{x^5}$ $+$ $\frac{1}{x^6}$

$= x^6 - 6x^4 + 15x^2 - 20 +$ $\frac{15}{x^2}$ $-$ $\frac{6}{x^4}$ $+$ $\frac{1}{x^6}$

$\\$

iv) $(1-3x)^7$

$= ^7 C_0 (3x)^0 - ^7 C_1 (3x)^1 + ^7 C_2 (3x)^2 - ^7 C_3 (3x)^3 +^7 C_4 (3x)^4 - ^7 C_5 (3x)^5 +^7 C_6 (3x)^6 \\ \\ { \hspace{12.0cm} - ^7 C_7 (3x)^7}$

$= 1 - 7 \times 3x + 21 \times 9x^2 - 35 \times 27 x^3 + 35 \times 81x^4 - 21 \times 24x^5 + 7 \times 729x^6 - 2187 x^7$

$= 1 - 21 x +189x^2 - 945x^3 + 2835x^4 - 5103x^5 + 5103x^6- 2187x^7$

$\\$

v) $\Big( ax -$ $\frac{b}{x}$ $\Big)^6$

$= ^6 C_0 (ax)^6 (\frac{b}{x})^0 - ^6 C_1 (ax)^5 (\frac{b}{x})^1 + ^6 C_2 (ax)^4 (\frac{b}{x})^2 - ^6 C_3 (ax)^3 (\frac{b}{x})^3 \\ \\ { \hspace{5.0cm} + ^6 C_4 (ax)^2 (\frac{b}{x})^4 - ^6 C_5 (ax)^1 (\frac{b}{x})^5 + ^6 C_6 (ax)^0 (\frac{b}{x})^6 }$

$= a^6x^6 - 6a^5x^5$ $( \frac{b}{x})$ $+ 15a^4x^4$ $( \frac{b^2}{x^2})$ $- 20 a^3b^3$ $( \frac{b^3}{x^3})$ $+ 15a^2x^2$ $( \frac{b^4}{x^4})$ $- 6ax$ $( \frac{b^5}{x^5})$ $+$ $( \frac{b^6}{x^6})$

$= a^6x^6 - 6a^5x^4b + 15a^4x^2b^2-20a^3b^3+15$ $(\frac{a^2b^4}{x^2} )$ $- 6$ $(\frac{ab^5}{x^4} )$ $+$ $( \frac{b^6}{x^6})$

$\\$

vi) $\Big($ $\sqrt{ \frac{x}{a}}$ $-$ $\sqrt{ \frac{a}{x}}$ $\Big)^6$

$= ^6 C_0 (\sqrt{\frac{x}{a}})^6 (\sqrt{\frac{a}{x}})^0 - ^6 C_1 (\sqrt{\frac{x}{a}})^5 (\sqrt{\frac{a}{x}})^1 +^6 C_2 (\sqrt{\frac{x}{a}})^4 (\sqrt{\frac{a}{x}})^2 \\ \\ { \hspace{0.0cm} - ^6 C_3 (\sqrt{\frac{x}{a}})^3 (\sqrt{\frac{a}{x}})^3 +^6 C_4 (\sqrt{\frac{x}{a}})^2 (\sqrt{\frac{a}{x}})^4 - ^6 C_5 (\sqrt{\frac{x}{a}})^1 (\sqrt{\frac{a}{x}})^5 + ^6 C_6 (\sqrt{\frac{x}{a}})^0 (\sqrt{\frac{a}{x}})^6}$

$=$ $( \frac{x^3}{a^3} )$ $- 6$ $( \frac{x^2}{a^2} )$ $+ 15$ $( \frac{x}{a} )$ $- 20 + 15$ $( \frac{a}{x} )$ $- 6$ $( \frac{a^2}{x^2} )$ $+$ $( \frac{a^3}{x^3} )$

$\\$

vii) $( \sqrt[3]{x} - \sqrt[3]{a} )^6$

$= ^6 C_0 (\sqrt[3]{x})^6 (\sqrt[3]{a})^0 - ^6 C_1 (\sqrt[3]{x})^5 (\sqrt[3]{a})^1 + ^6 C_2 (\sqrt[3]{x})^4 (\sqrt[3]{a})^2 - ^6 C_3 (\sqrt[3]{x})^3 (\sqrt[3]{a})^3 \\ \\ { \hspace{3.0cm} + ^6 C_4 (\sqrt[3]{x})^2 (\sqrt[3]{a})^4 - ^6 C_5 (\sqrt[3]{x})^1 (\sqrt[3]{a})^5 + ^6 C_0 (\sqrt[3]{x})^0 (\sqrt[3]{a})^6 }$

$= x^2 - 6 x^{\frac{5}{3}} a^{\frac{1}{3}} + 15 x^{\frac{4}{3}} a^{\frac{2}{3}} - 20xa + 15 x^{\frac{2}{3}} a^{\frac{4}{3}} - 6 x^{\frac{1}{3}} a^{\frac{5}{3}} + a^2$

$\\$

viii) $( 1 + 2x - 3x^2)^5 = [ (1+2x) - 3x^2 ]^5$

$= ^5 C_0 (1+2x)^5 (3x^2)^0 - ^5 C_1 (1+2x)^5 (3x^2)^1 +^5 C_2 (1+2x)^4 (3x^2)^2 \\ \\ { \hspace{3.0cm} - ^5 C_3 (1+2x)^3 (3x^2)^3 +^5 C_4 (1+2x)^2 (3x^2)^4 - ^5 C_5 (1+2x)^1 (3x^2)^5}$

$= (1+2x)^5 - 5 (1+2x)^4 \times (3x^2) + 10 (1+2x)^3 \times 9x^4 \\ \\ { \hspace{1.0cm}- 10 (1+2x)^2 \times 27x^6 + 5 (1+2x) \times 81x^8 - 243 x^{10} }$    … … … i)

Now we calculate the following:

$(1+2x)^5 = ^5 C_0 (2x)^5 +^5 C_1 (2x)^4 +^5 C_2 (2x)^3 +^5 C_3 (2x)^2 +^5 C_4 (2x)^1 +^5 C_5 (2x)^0$

$= 32x^5 + 80x^4 + 80^3 + 40x^2 + 10x + 1$

$(1+2x)^4 = ^4 C_0 (2x)^4 +^4 C_1 (2x)^3 +^4 C_2 (2x)^2 +^4 C_3 (2x)^1 +^4 C_4 (2x)^0$

$= 16x^4 + 32^3 + 24x^2 + 8x + 1$

$(1+2x)^3 = ^3 C_0 (2x)^3 +^3 C_1 (2x)^2 +^3 C_2 (2x)^1 +^3 C_3 (2x)^0$

$= 8^3 + 12x^2 + 6x + 1$

$(1+2x)^2 = ^2 C_0 (2x)^2 +^2 C_1 (2x)^1 +^2 C_2 (2x)^0$

$= 4x^2 + 4x + 1$

Substituting in i) we get

$= [32x^5 + 80x^4 + 80^3 + 40x^2 + 10x + 1] - 5 [16x^4 + 32^3 + 24x^2 + 8x + 1] \times (3x^2) \\ \\ { \hspace{0.0cm}+ 10 [8^3 + 12x^2 + 6x + 1] \times 9x^4 - 10 [4x^2 + 4x + 1] \times 27x^6 + 5 (1+2x) \times 81x^8 - 243 x^{10} }$

$= 1+10x+25x^2-40x^3-190x^4+92x^5+570x^6-360x^7-675x^8+810x^9-243x^{10}$

$\\$

ix) $\Big( x+1-$ $\frac{1}{x}$ $\Big)^3$

$= ^3 C_0 (x+1)^3 ( \frac{1}{x} )^0 - ^3 C_1 (x+1)^2 ( \frac{1}{x} )^1 + ^3 C_2 (x+1)^1 ( \frac{1}{x} )^2 - ^3 C_3 (x+1)^0 ( \frac{1}{x} )^3$

$= ( x+1)^3 - 3 \Big[$ $\frac{(x+1)^2}{x}$ $\Big]+ 3 \Big[$ $\frac{(x+1)}{x^2}$ $\Big] -$ $\frac{1}{x^3}$

$= [^3 C_0 x^3 + ^3 C_1 x^2 + ^3 C_2 x^1 + ^3 C_3 x^0 ] - 3 \Big[$ $\frac{x^2 + 1 + 2x}{x}$ $\Big] + 3 \Big[$ $\frac{(x+1)}{x^2}$ $\Big] -$ $\frac{1}{x^3}$

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

$= x^3 + 3x^2 -5 +$ $\frac{3}{x^2}$ $-$ $\frac{1}{x^3}$

$\\$

x) $( 1 - 2x + 3x^2)^3$

$= ^3 C_0 (1-2x)^3 (3x^2)^0 + ^3 C_1 (1-2x)^2 (3x^2)^1 + ^3 C_2 (1-2x)^1 (3x^2)^2 \\ \\ {\hspace{10.0cm}+ ^3 C_0 (1-2x)^0 (3x^2)^1 }$

$= (1-2x)^3 + 9x^2 ( 1- 2x)^2 + 27x^4 (1-2x) + 27x^6$

$= 1 - 6x + 21 x^2 - 44x^3 - 54x^5 + 27x^6$

$\\$

Question 2: Evaluate the following:

i) $( \sqrt{x+1} + \sqrt{x-1})^6 + ( \sqrt{x+1} - \sqrt{x-1})^6$

ii) $( x + \sqrt{x^2-1})^6 + ( x - \sqrt{x^2-1})^6$     iii) $(1+2\sqrt{x})^5-(1-2\sqrt{x})^5$

iv) $(\sqrt{2}+1)^6 + (\sqrt{2}-1)^6$      v) $(3 + \sqrt{2})^5 - (3 - \sqrt{2})^5$

vi) $(2 + \sqrt{3})^7 + (2 - \sqrt{3})^7$      vii) $(\sqrt{3} + 1)^5 - (\sqrt{3} - 1)^5$

viii) $(0.99)^5 + ( 1.01)^5$       ix) $(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6$

x) $\{a^2 + \sqrt{a^2-1} \}^4 + \{a^2 - \sqrt{a^2-1} \}^4$

i) $( \sqrt{x+1} + \sqrt{x-1})^6 + ( \sqrt{x+1} - \sqrt{x-1})^6$

$= 2 [ ^6 C_0 (\sqrt{x+1})^6 (\sqrt{x-1})^0 + ^6 C_2 (\sqrt{x+1})^4 (\sqrt{x-1})^2 \\ \\ { \hspace{3.0cm} + ^6 C_4 (\sqrt{x+1})^2 (\sqrt{x-1})^4 + ^6 C_6 (\sqrt{x+1})^0 (\sqrt{x-1})^4 ] }$

$= 2 [ (x+1)^3 + 15(x+1)^2 (x-1) + 15(x+1)(x-1)^2 + ( x-1)^3]$

$= 2 [ x^3 + 1 + 3x + 3x^2 + 15 ( x^2 + 2x+1)(x-1) + 15(x+1)^2(x^2+1-2x) \\ \\ { \hspace{10.0cm}+ x^3 - 1 + 3x -3x^2 ] }$

$= 2 [ 2x^3 + 6x + 15x^3 - 15x^2 +30x^2 - 30x + 15x - 15 + 15x^3 + 15x^2 \\ \\ { \hspace{10.0cm}- 30x^2 -30x + 15x+15] }$

$= 2 [ 32x^3 - 24x]$

$= 16x[4x^3 - 3]$

$\\$

ii) $( x + \sqrt{x^2-1})^6 + ( x - \sqrt{x^2-1})^6$

$= 2 [ ^6 C_0 x^6 (\sqrt{x^2-1})^0 + ^6 C_2 x^4 (\sqrt{x^2-1})^2 + ^6 C_4 x^2 (\sqrt{x^2-1})^4 + ^6 C_6 x^0 (\sqrt{x^2-1})^6 ]$

$= 2 [ x^6 + 15x^4 (x^2 -1)+ 15x^2 (x^2 -1)^2 + (x^2 -1)^3 ]$

$= 2 [ x^6 + 15x^6 - 15x^4 + 15x^2 (x^4 - 2x^2+1) + ( x^6 - 1 + 3x^2 - 3x^4) ]$

$= 2 [ x^6 + 15x^6 - 15x^4 + 15x^6-30x^4+15x^2+x^6 - 1 + 3x^2 - 3x^4) ]$

$= 64x^6 - 96x^4 + 36x^2 - 2$

$\\$

iii) $(1+2\sqrt{x})^5-(1-2\sqrt{x})^5$

$= 2 [ ^5 C_0 (2\sqrt{x})^0 + ^5 C_2 (2\sqrt{x})^2 + ^5 C_4 (2\sqrt{x})^4 ]$

$= 2 [ 1 + 10 \times 4x + 5 \times 16x^2]$

$= 2 ( 1 + 40x + 80x^2]$

$\\$

iv) $(\sqrt{2}+1)^6 + (\sqrt{2}-1)^6$

$= 2 [ ^6 C_0 (\sqrt{2})^6 + ^6 C_2 (\sqrt{2})^4 + ^6 C_4 (\sqrt{2})^2 + ^6 C_6 (\sqrt{2})^0 ]$

$= 2 [ 8 + 15 \times 4 + 15 \times 2 + 1]$

$= 2 \times 99$

$= 198$

$\\$

v) $(3 + \sqrt{2})^5 - (3 - \sqrt{2})^5$

$= 2 [ ^5 C_1 ( 3)^4 (\sqrt{2})^1 + ^5 C_3 ( 3)^2 (\sqrt{2})^3 + ^5 C_5 ( 3)^0 (\sqrt{2})^4 ]$

$= 2 [ 5 \times 81 \times \sqrt{2} + 10 \times 9 \times 2\sqrt{2} + 4 \sqrt{2} ]$

$= 2\sqrt{2} (405 + 180 + 4 )$

$= 1178 \sqrt{2}$

$\\$

vi) $(2 + \sqrt{3})^7 + (2 - \sqrt{3})^7$

$= 2 [ ^7 C_0 (2)^7(\sqrt{3})^0 + ^7 C_2 (2)^5(\sqrt{3})^2 + ^7 C_4 (2)^3(\sqrt{3})^4 + ^7 C_5 (2)^1(\sqrt{3})^6 ]$

$= 2 [ 128 + 21\times 32 \times 3 + 35 \times 8\times 9 + 7\times 2 \times 27 ]$

$=2 ( 128 + 2016 + 2520 + 378]$

$= 10084$

$\\$

vii) $(\sqrt{3} + 1)^5 - (\sqrt{3} - 1)^5$

$= 2 [ ^5 C_1 (\sqrt{3})^4 + ^5 C_3 (\sqrt{3})^2 + ^5 C_5 (\sqrt{3})^0 ]$

$= 2 [ 5 \times 9 + 10 \times 3 + 1 ]$

$= 2 \times 76$

$= 152$

$\\$

viii) $(0.99)^5 + ( 1.01)^5$

$= ( 1 - 0.01)^5 + ( 1+0.01)^5$

$= 2 [ ^5 C_0 (0.01)^0 + ^5 C_2 (0.01)^2 + ^5 C_4 (0.01)^4 ]$

$= 2 [ 1 + 10 \times 0.0001 + 5 \times 0.00000001 ]$

$= 2 \times 1.00100005$

$= 2.0020001$

$\\$

ix) $(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6$

$= 2 [ ^6 C_1 (\sqrt{3})^5 (\sqrt{2})^1 + ^6 C_3 (\sqrt{3})^3 (\sqrt{2})^3 + ^6 C_5 (\sqrt{3})^1 (\sqrt{2})^5 ]$

$= 2 [ 6 \times 9 \sqrt{3} \times \sqrt{2} + 20 \times 3 \sqrt{3} \times 2\sqrt{2} + 6 \times \sqrt{3} \times 4\sqrt{2}]$

$= 2 [ \sqrt{6}(54+120+24)]$

$= 396\sqrt{6}$

$\\$

x) $\{a^2 + \sqrt{a^2-1} \}^4 + \{a^2 - \sqrt{a^2-1} \}^4$

$= 2 [ ^4 C_0 (a^2)^4 (\sqrt{a^2-1})^0 + ^4 C_2 (a^2)^2 (\sqrt{a^2-1})^2 + ^4 C_4 (a^2)^0 (\sqrt{a^2-1})^4 ]$

$= 2[ a^8 + 6a^4(a^2-1) + (a^2-1)^2]$

$= 2[a^8+6a^6-6a^4+a^4+1-2a^2]$

$= 2a^2 + 12a^6 - 10a^4 - 4a^2 + 2$

$\\$

Question 3: Find $(a+b)^4 - ( a-b)^4$. Hence, evaluate $(\sqrt{3}+\sqrt{2})^4 - ( \sqrt{3}-\sqrt{2})^4$.

$(a+b)^4 - ( a-b)^4 = 2 [^4 C_1 a^3 b^1 + ^4 C_3 a^1 b^3 ] = 2 [ 4a^3b+4ab^3] = 8ab(a^2+b^2)$

Now we see, $a = \sqrt{3}$ and $b = \sqrt{2}$

In $(\sqrt{3}+\sqrt{2})^4 - ( \sqrt{3}-\sqrt{2})^4$

$= 8 \sqrt{3} \sqrt{2} [ (\sqrt{3})^2 + (\sqrt{2})^2]$

$= 40 \sqrt{6}$

$\\$

Question 4: Find $(x+1)^6 + (x-1)^6$. Hence, evaluate $( \sqrt{2}+1)^6+(\sqrt{2}-1)^6$

$(x+1)^6 + (x-1)^6 = 2[ ^6 C_0 x^6 +^6 C_2 x^4 +^6 C_4 x^2 +^6 C_6 x^0 ]$

$= 2 [ x^6 + 15x^4+15x^2+1]$

We see, $x = \sqrt{2}$

$( \sqrt{2}+1)^6+(\sqrt{2}-1)^6$

$= 2 [ (\sqrt{2})^6+ 15(\sqrt{2})^4 + 15 ( \sqrt{2})^2 + 1]$

$= 2 [ 8 + 60 + 30 + 1]$

$= 198$

$\\$

Question 5: Using binomial theorem evaluate each of the following:

i) $(96)^3$     ii) $(102)^5$     iii) $(101)^4$     iv)  $(98)^5$

i) $(96)^3 = (100-4)^3$

$= ^3 C_0 (100)^3 ( -4)^0 +^3 C_1 (100)^2 ( -4)^1 +^3 C_2 (100)^1 ( -4)^2 +^3 C_3 (100)^0 ( -4)^3$

$= 100^3 - 3 \times 100^2 \times 4 + 3 \times 100 \times 4^2 - 4^3$

$= 1000000 - 120000 + 4800-64$

$= 1004800-120064$

$= 884736$

ii) $(102)^5 = ( 100+2)^5$

$= ^5 C_0 (100)^5 ( 2)^0 + ^5 C_1 (100)^4 ( 2)^1 + ^5 C_2 (100)^3 ( 2)^2 + ^5 C_3 (100)^2 ( 2)^3 \\ \\ { \hspace{6.0cm}+ ^5 C_4 (100)^1 ( 2)^4 + ^5 C_5 (100)^0 ( 2)^5}$

$= 100^5 + 5 \times 100^4 \times 2 + 10 \times 100^3 \times 2^2 + 10 \times 100^2 \times 2^3 + 5 \times 100 \times 2^4 + 2^5$

$= 10000000000 + 1000000000 + 40000000+ 800000+8000+32$

$= 11040808032$

iii) $(101)^4 = (100+1)^4$

$= ^4 C_0 (100)^4 + ^4 C_1 (100)^3 + ^4 C_2 (100)^2 + ^4 C_3 (100)^1 + ^4 C_4 (100)^0$

$= 100^4 + 4 \times 100^3 + 6 \times 100^2 + 400 + 1$

$= 100000000 + 4000000+60000+400+1$

$= 104060401$

iv)  $(98)^5 = ( 100 - 2)^5$

$= ^5 C_0 (100)^5 (-2)^0 + ^5 C_1 (100)^4 (-2)^1 + ^5 C_2 (100)^3 (-2)^2 + ^5 C_3 (100)^2 (-2)^3 + ^5 C_4 (100)^1 (-2)^4 + ^5 C_5 (100)^0 (-2)^5$

$= 100^5 - 10 \times 100^4 + 40 \times 100^3 - 80 \times 100^2 + 80 \times 100 - 32$

$= 10000000000 - 1000000000+40000000-800000+8000-32$

$= 10040008000-1000800032$

$= 9039207968$

$\\$

Question 6: Using binomial theorem prove that $2^{3n}-7n-1$ is divisible by $49$, where $n \in N$

$2^{3n}-7n-1$

$= 8^n - 7n -1$

$= (1+7)^n - 7n - 1$

$= [ ^n C_0 7^0+ ^n C_1 7^1+ ^n C_2 7^2+ \ldots + ^n C_n 7^n ] - 7n - 1$

$= 1+7n + [ ^n C_2 7^2+ \ldots + ^n C_n 7^n ] - 7n - 1$

$= [ ^n C_2 7^2+ \ldots + ^n C_n 7^n ]$

$= 49 [ ^n C_2 + \ldots + ^n C_n 7^{n-2} ]$

Hence $2^{3n}-7n-1$ is divisible by $49$, where $n \in N$

$\\$

Question 7: Using binomial theorem prove that $3^{2n+2}-8n-9$ is divisible by $64, n \in N$

$3^{2n+2}-8n-9$

$= 9^{n+1} - 8n - 9$

$= (1+8)^{n+1} - 8n - 9$

$= [ ^{n+1} C_0 8^0 + ^{n+1} C_1 8^1 + ^{n+1} C_2 8^2 + \ldots + ^{n+1} C_{n+1} 8^{n+1} ] - 8n - 9$

$= 1 + 8(n+1) + [ ^{n+1} C_2 8^2 + \ldots + ^{n+1} C_{n+1} 8^{n+1} ] - 8n - 9$

$= [ ^{n+1} C_2 8^2 + \ldots + ^{n+1} C_{n+1} 8^{n+1} ]$

$= 64 [ ^{n+1} C_2 + \ldots + ^{n+1} C_{n+1} 8^{n-1} ]$

Hence $3^{2n+2}-8n-9$ is divisible by $64, n \in N$

$\\$

Question 8: If $n$ is a positive integer, prove that $3^{3n}-26n-1$ is divisible by $676$.

$3^{3n}-26n-1$

$= 27^n - 26n - 1$

$= (1+26)^n - 26n - 1$

$= [ ^{n} C_0 (26)^0 + ^{n} C_1 (26)^1 + ^{n} C_2 (26)^2 + \ldots + ^{n} C_{n+1} (26)^{n} ]- 26n - 1$

$= 1+26n + [^{n} C_2 (26)^2 + \ldots + ^{n} C_{n+1} (26)^{n} ]- 26n - 1$

$= [^{n} C_2 (26)^2 + \ldots + ^{n} C_{n+1} (26)^{n} ]$

$= 676 [^{n} C_2 + \ldots + ^{n} C_{n+1} (26)^{n-2} ]$

Hence $3^{3n}-26n-1$ is divisible by $676$.

$\\$

Question 9: Using binomial theorem, indicate which is larger $(1.1)^{10000}$ or $10000$.

$(1.1)^{10000}$

$= (1+0.1)^{10000}$

$= ^{10000} C_0 (0.1)^0 + ^{10000} C_1 (0.1)^1 +^{10000} C_2 (0.1)^2 + \ldots + ^{10000} C_{10000} (0.1)^{10000}$

$= 1 + 10000(0.1) + \text{other positive terms}$

$= 10001 + \text{other positive terms}$

Therefore $(1.1)^{10000}$ is greater than $10000$ .

$\\$

Question 10: Using binomial theorem, determine which is smaller $(1.2)^{4000}$ or $800$.

$(1.2)^{4000}$

$= (1+0.2)^{4000}$

$= ^{4000} C_0 (0.2)^0 + ^{4000} C_1 (0.2)^1 +^{4000} C_2 (0.2)^2 + \ldots + ^{4000} C_{4000} (0.1)^{4000}$

$= 1 + 4000(0.2) + \text{other positive terms}$

$= 801 + \text{other positive terms}$

Therefore $800$ is the smaller term.

$\\$

Question 11: Find the value of $(1.01)^{10}+(1-0.01)^{10}$ correct to $7$ decimal places.

$(1.01)^{10}+(1-0.01)^{10}$

$= (1+ 0.01)^{10}+(1-0.01)^{10}$

$= 2 [ ^{10} C_0 (0.01)^0 + ^{10} C_2 (0.01)^2 +^{10} C_4 (0.01)^4 +^{10} C_6 (0.01)^6 +^{10} C_8 (0.01)^8 +^{10} C_{10} (0.01)^{10} ]$

$= 2 [ 1 + 45 ( 0.0001) + 210 ( 0.0000001) + \ldots ]$

$= 2.0090042 + \ldots$

Hence $(1.01)^{10}+(1-0.01)^{10}$ correct to $7$ decimal places is $2.0090042$

$\\$

Question 12: Show that $2^{4n+4}- 15n - 16$, where $n \in N$ is divisible by $225$.

$2^{4n+4}- 15n - 16$

$= {16}^{n+1}- 15n - 16$

$= (1+15)^{n+1}- 15n - 16$

$= [^{n+1} C_0 (15)^0 +^{n+1} C_1 (15)^1 +^{n+1} C_2 (15)^2 + \ldots + ^{n+1} C_{n+1} (15)^{n+1} ]- 15n - 16$

$= 1 + 15( n+1) + [^{n+1} C_2 (15)^2 + \ldots + ^{n+1} C_{n+1} (15)^{n+1} ] - 15n - 16$

$= [^{n+1} C_2 (15)^2 + \ldots + ^{n+1} C_{n+1} (15)^{n+1} ]$

$= 225 [^{n+1} C_2 + \ldots + ^{n+1} C_{n+1} (15)^{n-1} ]$

Hence $2^{4n+4}- 15n - 16$, where $n \in N$ is divisible by $225$.