Answer :

By using Binomial Theorem, the expression $(1- 2 x)^{5}$ can be expanded as

$(1-2 x)^{5}$

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

$=1-5(2 x)+10(4 x^{2})-10(8 x^{3})+5(16 x^{4})-(32 x^{5})$

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

Answer :

By using Binomial Theorem, the expression $(\dfrac{2}{x}-\dfrac{x}{2})^{5}$ can be expanded as

$ \begin{aligned} (\dfrac{2}{x}-\dfrac{x}{2})^{5}= & { }^{5} C_0(\dfrac{2}{x})^{5}-{ }^{5} C_1(\dfrac{2}{x})^{4}(\dfrac{x}{2})+{ }^{5} C_2(\dfrac{2}{x})^{3}(\dfrac{x}{2})^{2} \\ & -{ }^{5} C_3(\dfrac{2}{x})^{2}(\dfrac{x}{2})^{3}+{ }^{5} C_4(\dfrac{2}{x})(\dfrac{x}{2})^{4}-{ }^{5} C_5(\dfrac{x}{2})^{5} \\ = & \dfrac{32}{x^{5}}-5(\dfrac{16}{x^{4}})(\dfrac{x}{2})+10(\dfrac{8}{x^{3}})(\dfrac{x^{2}}{4})-10(\dfrac{4}{x^{2}})(\dfrac{x^{3}}{8})+5(\dfrac{2}{x})(\dfrac{x^{4}}{16})-\dfrac{x^{5}}{32} \\ = & \dfrac{32}{x^{5}}-\dfrac{40}{x^{3}}+\dfrac{20}{x}-5 x+\dfrac{5}{8} x^{3}-\dfrac{x^{5}}{32} \end{aligned} $

Answer :

By using Binomial Theorem, the expression (2x- 3$)^{6}$ can be expanded as

$ \begin{aligned} (2 x-3)^{6}= & { }^{6} C_0(2 x)^{6}-{ }^{6} C_1(2 x)^{5}(3)+{ }^{6} C_2(2 x)^{4}(3)^{2}-{ }^{6} C_3(2 x)^{3}(3)^{3} \\ & +{ }^{6} C_4(2 x)^{2}(3)^{4}-{ }^{6} C_5(2 x)(3)^{5}+{ }^{6} C_6(3)^{6} \\ = & 64 x^{6}-6(32 x^{5})(3)+15(16 x^{4})(9)-20(8 x^{3})(27) \\ & +15(4 x^{2})(81)-6(2 x)(243)+729 \\ = & 64 x^{6}-576 x^{5}+2160 x^{4}-4320 x^{3}+4860 x^{2}-2916 x+729 \end{aligned} $

Answer :

By using Binomial Theorem, the expression $(\dfrac{x}{3}+\dfrac{1}{x})^{5}$ can be expanded as

$ \begin{aligned} (\dfrac{x}{3}+\dfrac{1}{x})^{5}= & { }^{5} C_0(\dfrac{x}{3})^{5}+{ }^{5} C_1(\dfrac{x}{3})^{4}(\dfrac{1}{x})+{ }^{5} C_2(\dfrac{x}{3})^{3}(\dfrac{1}{x})^{2} \\ & +{ }^{5} C_3(\dfrac{x}{3})^{2}(\dfrac{1}{x})^{3}+{ }^{5} C_4(\dfrac{x}{3})(\dfrac{1}{x})^{4}+{ }^{5} C_5(\dfrac{1}{x})^{5} \\ = & \dfrac{x^{5}}{243}+5(\dfrac{x^{4}}{81})(\dfrac{1}{x})+10(\dfrac{x^{3}}{27})(\dfrac{1}{x^{2}})+10(\dfrac{x^{2}}{9})(\dfrac{1}{x^{3}})+5(\dfrac{x}{3})(\dfrac{1}{x^{4}})+\dfrac{1}{x^{5}} \\ = & \dfrac{x^{5}}{243}+\dfrac{5 x^{3}}{81}+\dfrac{10 x}{27}+\dfrac{10}{9 x}+\dfrac{5}{3 x^{3}}+\dfrac{1}{x^{5}} \end{aligned} $

Answer :

By using Binomial Theorem, the expression $(x+\dfrac{1}{x})^{6}$ can be expanded as

$ \begin{aligned} (x+\dfrac{1}{x})^{6}= & { }^{6} C_0(x)^{6}+{ }^{6} C_1(x)^{5}(\dfrac{1}{x})+{ }^{6} C_2(x)^{4}(\dfrac{1}{x})^{2} \\ & +{ }^{6} C_3(x)^{3}(\dfrac{1}{x})^{3}+{ }^{6} C_4(x)^{2}(\dfrac{1}{x})^{4}+{ }^{6} C_5(x)(\dfrac{1}{x})^{5}+{ }^{6} C_6(\dfrac{1}{x})^{6} \\ = & x^{6}+6(x)^{5}(\dfrac{1}{x})+15(x)^{4}(\dfrac{1}{x^{2}})+20(x)^{3}(\dfrac{1}{x^{3}})+15(x)^{2}(\dfrac{1}{x^{4}})+6(x)(\dfrac{1}{x^{5}})+\dfrac{1}{x^{6}} \\ = & x^{6}+6 x^{4}+15 x^{2}+20+\dfrac{15}{x^{2}}+\dfrac{6}{x^{4}}+\dfrac{1}{x^{6}} \end{aligned} $

Using binomial theorem, evaluate each of the following:

Answer :

96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.

It can be written that, $96=100$- 4

$ \begin{aligned} \therefore(96)^{3} & =(100-4)^{3} \\ & ={ }^{3} C_0(100)^{3}-{ }^{3} C_1(100)^{2}(4)+{ }^{3} C_2(100)(4)^{2}-{ }^{3} C_3(4)^{3} \\ & =(100)^{3}-3(100)^{2}(4)+3(100)(4)^{2}-(4)^{3} \\ & =1000000-120000+4800-64 \\ & =884736 \end{aligned} $

Answer :

102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, $102=100+2$

$ \begin{aligned} \therefore(102)^{5}= & (100+2)^{5} \\ = & { }^{5} C_0(100)^{5}+{ }^{5} C_1(100)^{4}(2)+{ }^{5} C_2(100)^{3}(2)^{2}+{ }^{5} C_3(100)^{2}(2)^{3} \\ & +{ }^{5} C_4(100)(2)^{4}+{ }^{5} C_5(2)^{5} \\ = & (100)^{5}+5(100)^{4}(2)+10(100)^{3}(2)^{2}+10(100)^{2}(2)^{3}+5(100)(2)^{4}+(2)^{5} \\ = & 10000000000+1000000000+40000000+800000+8000+32 \\ = & 11040808032 \end{aligned} $

Answer :

101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, $101=100+1$

$ \begin{aligned} \therefore(101)^{4} & =(100+1)^{4} \\ & ={ }^{4} C_0(100)^{4}+{ }^{4} C_1(100)^{3}(1)+{ }^{4} C_2(100)^{2}(1)^{2}+{ }^{4} C_3(100)(1)^{3}+{ }^{4} C_4(1)^{4} \\ & =(100)^{4}+4(100)^{3}+6(100)^{2}+4(100)+(1)^{4} \\ & =100000000+4000000+60000+400+1 \\ & =104060401 \end{aligned} $

Answer :

99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, $99=100$- 1

$ \begin{aligned} \therefore(99)^{5}= & (100-1)^{5} \\ = & { }^{5} C_0(100)^{5}-{ }^{5} C_1(100)^{4}(1)+{ }^{5} C_2(100)^{3}(1)^{2}-{ }^{5} C_3(100)^{2}(1)^{3} \\ & +{ }^{5} C_4(100)(1)^{4}-{ }^{5} C_5(1)^{5} \\ = & (100)^{5}-5(100)^{4}+10(100)^{3}-10(100)^{2}+5(100)-1 \\ = & 10000000000-500000000+10000000-100000+500-1 \\ = & 10010000500-500100001 \\ = & 9509900499 \end{aligned} $

Answer :

By splitting 1.1 and then applying Binomial Theorem, the first few terms of $(1.1)^{10000}$ can be obtained as

$ \begin{aligned} (1.1)^{10000} & =(1+0.1)^{10000} \\ & ={ }^{10000} C_0+{ }^{10000} C_1(1.1)+\text{ Other positive terms } \\ & =1+10000 \times 1.1+\text{ Other positive terms } \\ & =1+11000+\text{ Other positive terms } \\ & >1000 \end{aligned} $

Hence, $(1.1)^{10000}>1000$

Answer :

Using Binomial Theorem, the expressions, $(a+b)^{4}$ and $(a-b)^{4}$, can be expanded as

$ \begin{aligned} & (a+b)^{4}={ }^{4} C_0 a^{4}+{ }^{4} C_1 a^{3} b+{ }^{4} C_2 a^{2} b^{2}+{ }^{4} C_3 ab^{3}+{ }^{4} C_4 b^{4} \\ & (a-b)^{4}={ }^{4} C_0 a^{4}-{ }^{4} C_1 a^{3} b+{ }^{4} C_2 a^{2} b^{2}-{ }^{4} C_3 a b^{3}+{ }^{4} C_4 b^{4} \\ & \begin{aligned} \therefore(a+b)^{4}-(a-b)^{4}= & { }^{4} C_0 a^{4}+{ }^{4} C_1 a^{3} b+{ }^{4} C_2 a^{2} b^{2}+{ }^{4} C_3 a b^{3}+{ }^{4} C_4 b^{4} \\ & -[{ }^{4} C_0 a^{4}-{ }^{4} C_1 a^{3} b+{ }^{4} C_2 a^{2} b^{2}-{ }^{4} C_3 a b^{3}+{ }^{4} C_4 b^{4}] \\ = & 2({ }^{4} C_1 a^{3} b+{ }^{4} C_3 a b^{3})=2(4 a^{3} b+4 a b^{3}) \\ = & 8 ab(a^{2}+b^{2}) \end{aligned} \end{aligned} $

By putting $a=\sqrt{3}$ and $b=\sqrt{2}$, we obtain

$ \begin{aligned} (\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4} & =8(\sqrt{3})(\sqrt{2})\{(\sqrt{3})^{2}+(\sqrt{2})^{2}\} \\ & =8(\sqrt{6})\{3+2\}=40 \sqrt{6} \end{aligned} $

Answer :

Using Binomial Theorem, the expressions, $(x+1)^{6}$ and $(x - 1)^{6}$, can be expanded as

$ \begin{gathered} (x+1)^{6}={ }^{6} C_0 x^{6}+{ }^{6} C_1 x^{3}+{ }^{6} C_2 x^{4}+{ }^{6} C_3 x^{3}+{ }^{6} C_4 x^{2}+{ }^{6} C_5 x+{ }^{6} C_6 \\ (x-1)^{6}={ }^{6} C_0 x^{6}-{ }^{6} C_1 x^{5}+{ }^{6} C_2 x^{4}-{ }^{6} C_3 x^{3}+{ }^{6} C_4 x^{2}-{ }^{6} C_5 x+{ }^{6} C_6 \\ \therefore(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] \\ =2[x^{6}+15 x^{4}+15 x^{2}+1] \end{gathered} $

By putting $x=\sqrt{2}$, we obtain

$ \begin{aligned} (\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6} & =2[(\sqrt{2})^{6}+15(\sqrt{2})^{4}+15(\sqrt{2})^{2}+1] \\ & =2(8+15 \times 4+15 \times 2+1) \\ & =2(8+60+30+1) \\ & =2(99)=198 \end{aligned} $

Answer :

In order to show that $9^{n+1}-8 n-9$ is divisible by 64 , it has to be proved that,

$9^{n+1}-8 n-9=64 k$, where $k$ is some natural number

By Binomial Theorem,

$(1+a)^{m}={ }^{m} C_0+{ }^{m} C_1 a+{ }^{m} C_2 a^{2}+\ldots+{ }^{m} C_m a^{m}$

For $a=8$ and $m=n+1$, we obtain

$(1+8)^{n+1}={ }^{n+1} C_0+{ }^{n+1} C_1(8)+{ }^{n+1} C_2(8)^{2}+\ldots+{ }^{n+1} C _{n+1}(8)^{n+1}$

$\Rightarrow 9^{n+1}=1+(n+1)(8)+8^{2}[{ }^{n+1} C_2+{ }^{n+1} C_3 \times 8+\ldots+{ }^{n+1} C _{n+1}(8)^{n-1}]$

$\Rightarrow 9^{n+1}=9+8 n+64[{ }^{n+1} C_2+{ }^{n+1} C_3 \times 8+\ldots+{ }^{n+1} C _{n+1}(8)^{n-1}]$

$\Rightarrow 9^{n+1}-8 n-9=64 k$, where $k={ }^{n+1} C_2+{ }^{n+1} C_3 \times 8+\ldots+{ }^{n+1} C _{n+1}(8)^{n-1}$ is a natural number

Thus, $9^{n+1}-8 n-9$ is divisible by 64 , whenever nis a positive integer.

Answer : By Binomial Theorem,

$ \sum _{r=0}^{n}{ }^{n} C_r a^{n-r} b^{r}=(a+b)^{n} $

By putting $b=3$ and $a=1$ in the above equation, we obtain

$ \begin{aligned} & \sum _{r=0}^{n}{ }^{n} C_r(1)^{n-r}(3)^{r}=(1+3)^{n} \\ & \Rightarrow \sum _{r=0}^{n} 3^{r}{ }^{n} C_r=4^{n} \end{aligned} $

Hence, proved.