Answer :

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

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

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

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

Answer :

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

$ \begin{aligned} \left(\dfrac{2}{x}-\dfrac{x}{2}\right)^{5} \ = \ & { }^{5} C_0\left(\dfrac{2}{x}\right)^{5}-{ }^{5} C_1\left(\dfrac{2}{x}\right)^{4}\left(\dfrac{x}{2}\right)+{ }^{5} C_2\left(\dfrac{2}{x}\right)^{3}\left(\dfrac{x}{2}\right)^{2} -{ }^{5} C_3\left(\dfrac{2}{x}\right)^{2}\left(\dfrac{x}{2}\right)^{3}+{ }^{5} C_4\left(\dfrac{2}{x}\right)\left(\dfrac{x}{2}\right)^{4}-{ }^{5} C_5\left(\dfrac{x}{2}\right)^{5} \\ \\ \ = \ & \dfrac{32}{x^{5}}-5\left(\dfrac{16}{x^{4}}\right)\left(\dfrac{x}{2}\right)+10\left(\dfrac{8}{x^{3}}\right)\left(\dfrac{x^{2}}{4}\right)-10\left(\dfrac{4}{x^{2}}\right)\left(\dfrac{x^{3}}{8}\right)+5\left(\dfrac{2}{x}\right)\left(\dfrac{x^{4}}{16}\right)-\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 $\left(2x- 3\right)^{6}$ can be expanded as

$ \begin{aligned} \left(2 x-3\right)^{6} \ = \ & { }^{6} C_0\left(2 x\right)^{6}-{ }^{6} C_1\left(2 x\right)^{5}\left(3\right)+{ }^{6} C_2\left(2 x\right)^{4}\left(3\right)^{2}-{ }^{6} C_3\left(2 x\right)^{3}\left(3\right)^{3} +{ }^{6} C_4\left(2 x\right)^{2}\left(3\right)^{4}-{ }^{6} C_5\left(2 x\right)\left(3\right)^{5}+{ }^{6} C_6\left(3\right)^{6} \\ \\ \ = \ & 64 x^{6}-6\left(32 x^{5}\right)\left(3\right)+15\left(16 x^{4}\right)\left(9\right)-20\left(8 x^{3}\right)\left(27\right) +15\left(4 x^{2}\right)\left(81\right)-6\left(2 x\right)\left(243\right)+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 $\left(\dfrac{x}{3}+\dfrac{1}{x}\right)^{5}$ can be expanded as

$ \begin{aligned} \left(\dfrac{x}{3}+\dfrac{1}{x}\right)^{5} \ = \ & { }^{5} C_0\left(\dfrac{x}{3}\right)^{5}+{ }^{5} C_1\left(\dfrac{x}{3}\right)^{4}\left(\dfrac{1}{x}\right)+{ }^{5} C_2\left(\dfrac{x}{3}\right)^{3}\left(\dfrac{1}{x}\right)^{2} +{ }^{5} C_3\left(\dfrac{x}{3}\right)^{2}\left(\dfrac{1}{x}\right)^{3}+{ }^{5} C_4\left(\dfrac{x}{3}\right)\left(\dfrac{1}{x}\right)^{4}+{ }^{5} C_5\left(\dfrac{1}{x}\right)^{5} \\ \\ \ = \ & \dfrac{x^{5}}{243}+5\left(\dfrac{x^{4}}{81}\right)\left(\dfrac{1}{x}\right)+10\left(\dfrac{x^{3}}{27}\right)\left(\dfrac{1}{x^{2}}\right)+10\left(\dfrac{x^{2}}{9}\right)\left(\dfrac{1}{x^{3}}\right)+5\left(\dfrac{x}{3}\right)\left(\dfrac{1}{x^{4}}\right)+\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 $\left(x+\dfrac{1}{x}\right)^{6}$ can be expanded as

$ \begin{aligned} \left(x+\dfrac{1}{x}\right)^{6} \ = \ & { }^{6} C_0\left(x\right)^{6}+{ }^{6} C_1\left(x\right)^{5}\left(\dfrac{1}{x}\right)+{ }^{6} C_2\left(x\right)^{4}\left(\dfrac{1}{x}\right)^{2} +{ }^{6} C_3\left(x\right)^{3}\left(\dfrac{1}{x}\right)^{3}+{ }^{6} C_4\left(x\right)^{2}\left(\dfrac{1}{x}\right)^{4}+{ }^{6} C_5\left(x\right)\left(\dfrac{1}{x}\right)^{5}+{ }^{6} C_6\left(\dfrac{1}{x}\right)^{6} \\ \\ \ = \ & x^{6}+6\left(x\right)^{5}\left(\dfrac{1}{x}\right)+15\left(x\right)^{4}\left(\dfrac{1}{x^{2}}\right)+20\left(x\right)^{3}\left(\dfrac{1}{x^{3}}\right)+15\left(x\right)^{2}\left(\dfrac{1}{x^{4}}\right)+6\left(x\right)\left(\dfrac{1}{x^{5}}\right)+\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 \ \ \left(96\right)^{3} & =\left(100-4\right)^{3} \\ \\ & ={ }^{3} C_0\left(100\right)^{3}-{ }^{3} C_1\left(100\right)^{2}\left(4\right)+{ }^{3} C_2\left(100\right)\left(4\right)^{2}-{ }^{3} C_3\left(4\right)^{3} \\ \\ & =\left(100\right)^{3}-3\left(100\right)^{2}\left(4\right)+3\left(100\right)\left(4\right)^{2}-\left(4\right)^{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 \ \ \left(102\right)^{5} \ = \ & \left(100+2\right)^{5} \\ \\ \ = \ & { }^{5} C_0\left(100\right)^{5}+{ }^{5} C_1\left(100\right)^{4}\left(2\right)+{ }^{5} C_2\left(100\right)^{3}\left(2\right)^{2}+{ }^{5} C_3\left(100\right)^{2}\left(2\right)^{3} +{ }^{5} C_4\left(100\right)\left(2\right)^{4}+{ }^{5} C_5\left(2\right)^{5} \\ \\ \ = \ & \left(100\right)^{5}+5\left(100\right)^{4}\left(2\right)+10\left(100\right)^{3}\left(2\right)^{2}+10\left(100\right)^{2}\left(2\right)^{3}+5\left(100\right)\left(2\right)^{4}+\left(2\right)^{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 \ \ \left(101\right)^{4} & =\left(100+1\right)^{4} \\ \\ & ={ }^{4} C_0\left(100\right)^{4}+{ }^{4} C_1\left(100\right)^{3}\left(1\right)+{ }^{4} C_2\left(100\right)^{2}\left(1\right)^{2}+{ }^{4} C_3\left(100\right)\left(1\right)^{3}+{ }^{4} C_4\left(1\right)^{4} \\ \\ & =\left(100\right)^{4}+4\left(100\right)^{3}+6\left(100\right)^{2}+4\left(100\right)+\left(1\right)^{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 \ \ \left(99\right)^{5} \ = \ & \left(100-1\right)^{5} \\ \\ \ = \ & { }^{5} C_0\left(100\right)^{5}-{ }^{5} C_1\left(100\right)^{4}\left(1\right)+{ }^{5} C_2\left(100\right)^{3}\left(1\right)^{2}-{ }^{5} C_3\left(100\right)^{2}\left(1\right)^{3} +{ }^{5} C_4\left(100\right)\left(1\right)^{4}-{ }^{5} C_5\left(1\right)^{5} \\ \\ \ = \ & \left(100\right)^{5}-5\left(100\right)^{4}+10\left(100\right)^{3}-10\left(100\right)^{2}+5\left(100\right)-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 $\left(1.1\right)^{10000}$ can be obtained as

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

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

Answer :

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

$ \begin{aligned} & \left(a+b\right)^{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} \\ \\ & \left(a-b\right)^{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 \ \ \left(a+b\right)^{4}-\left(a-b\right)^{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} -\big[{ }^{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}\big] \\ \\ \ = \ & 2\left({ }^{4} C_1 a^{3} b+{ }^{4} C_3 a b^{3}\right)=2\left(4 a^{3} b+4 a b^{3}\right) \\ \\ \ = \ & 8 ab\left(a^{2}+b^{2}\right) \end{aligned} \end{aligned} $

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

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

Answer :

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

$ \begin{aligned} & \left(x+1\right)^{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 \\ \\ & \left(x-1\right)^{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 \ \ \left(x+1\right)^{6}+\left(x-1\right)^{6}=2\big[{ }^{6} C_0 x^{6}+{ }^{6} C_2 x^{4}+{ }^{6} C_4 x^{2}+{ }^{6} C_6\big] =2\big[x^{6}+15 x^{4}+15 x^{2}+1\big] \end{aligned} $

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

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

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

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

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

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

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

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

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

Answer :

By Binomial Theorem,

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

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

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

Hence, proved.