Chapter 7: Binomial Theorem

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

\[\begin{gathered} {\left( {1 - 2x} \right)^5} = {}^5{C_0}{\left( 1 \right)^5} - {}^5{C_1}{\left( 1 \right)^4}\left( {2x} \right) + {}^5{C_2}{\left( 1 \right)^3}{\left( {2x} \right)^2} - {}^5{C_3}{\left( 1 \right)^2}{\left( {2x} \right)^3} + {}^5{C_4}{\left( 1 \right)^1}{\left( {2x} \right)^4} \\ - {}^5{C_5}{\left( {2x} \right)^5} \\ = 1 - 5\left( {2x} \right) + 10\left( {4{x^2}} \right) - 10\left( {8{x^3}} \right) + 5\left( {16{x^4}} \right) - 32{x^5} \\ = 1 - 10x + 40{x^2} - 80{x^3} + 80{x^4} - 32{x^5} \\ \end{gathered}\]

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

\[\begin{gathered} {\left( {\frac{2}{x} - \frac{x}{2}} \right)^5} = {}^5{C_0}{\left( {\frac{2}{x}} \right)^5} - {}^5{C_1}{\left( {\frac{2}{x}} \right)^4}\left( {\frac{x}{2}} \right) + {}^5{C_2}{\left( {\frac{2}{x}} \right)^3}{\left( {\frac{x}{2}} \right)^2} - {}^5{C_3}{\left( {\frac{2}{x}} \right)^2}{\left( {\frac{x}{2}} \right)^3} + {}^5{C_4}{\left( {\frac{2}{x}} \right)^1}{\left( {\frac{x}{2}} \right)^4} \\ - {}^5{C_5}{\left( {\frac{x}{2}} \right)^5} \\ = \frac{{32}}{{{x^5}}} - 5\left( {\frac{{16}}{{{x^4}}}} \right)\left( {\frac{x}{2}} \right) + 10\left( {\frac{8}{{{x^3}}}} \right)\left( {\frac{{{x^2}}}{4}} \right) - 10\left( {\frac{4}{{{x^2}}}} \right)\left( {\frac{{{x^3}}}{8}} \right) + 5\left( {\frac{2}{x}} \right)\left( {\frac{{{x^4}}}{{16}}} \right) - \frac{{{x^5}}}{{32}} \\ = \frac{{32}}{{{x^5}}} - \frac{{40}}{{{x^3}}} + \frac{{20}}{x} - 5x + \frac{5}{8}{x^3} - \frac{{{x^5}}}{{32}} \\ \end{gathered}\]

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

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

Ans. By using Binomial Theorem, the expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$ can be expanded as

\[\begin{gathered} {\left( {\frac{x}{3} + \frac{1}{x}} \right)^5} = {}^5{C_0}{\left( {\frac{x}{3}} \right)^5} + {}^5{C_1}{\left( {\frac{x}{3}} \right)^4}\left( {\frac{1}{x}} \right) + {}^5{C_2}{\left( {\frac{x}{3}} \right)^3}{\left( {\frac{1}{x}} \right)^2} + {}^5{C_3}{\left( {\frac{x}{3}} \right)^2}{\left( {\frac{1}{x}} \right)^3} + {}^5{C_4}{\left( {\frac{x}{3}} \right)^1}{\left( {\frac{1}{x}} \right)^4} \\ + {}^5{C_5}{\left( {\frac{1}{x}} \right)^5} \\ = \frac{{{x^5}}}{{243}} + 5\left( {\frac{{{x^4}}}{{81}}} \right)\left( {\frac{1}{x}} \right) + 10\left( {\frac{{{x^3}}}{{27}}} \right)\left( {\frac{1}{{{x^2}}}} \right) + 10\left( {\frac{{{x^2}}}{9}} \right)\left( {\frac{1}{{{x^3}}}} \right) + 5\left( {\frac{x}{3}} \right)\left( {\frac{1}{{{x^4}}}} \right) + \frac{1}{{{x^5}}} \\ = \frac{{{x^5}}}{{243}} + \frac{{5{x^3}}}{{81}} + \frac{{10x}}{{27}} + \frac{{10}}{{9x}} + \frac{5}{{3{x^3}}} + \frac{1}{{{x^5}}} \\ \end{gathered} \]

Ans. By using Binomial Theorem, the expression ${\left( {x + \frac{1}{x}} \right)^6}$ can be expanded as

\[\begin{gathered} {\left( {x + \frac{1}{x}} \right)^6} = {}^6{C_0}{\left( x \right)^6} + {}^6{C_1}{\left( x \right)^5}\left( {\frac{1}{x}} \right) + {}^6{C_2}{\left( x \right)^4}{\left( {\frac{1}{x}} \right)^2} + {}^6{C_3}{\left( x \right)^3}{\left( {\frac{1}{x}} \right)^3} + {}^6{C_4}{\left( x \right)^2}{\left( {\frac{1}{x}} \right)^4} \\ + {}^6{C_5}\left( x \right){\left( {\frac{1}{x}} \right)^5} + {}^6{C_6}{\left( {\frac{1}{x}} \right)^6} \\ = {x^6} + 6\left( {{x^5}} \right)\left( {\frac{1}{x}} \right) + 15\left( {{x^4}} \right)\left( {\frac{1}{{{x^2}}}} \right) + 20\left( {{x^3}} \right)\left( {\frac{1}{{{x^3}}}} \right) + 15\left( {{x^2}} \right)\left( {\frac{1}{{{x^4}}}} \right) + 6\left( x \right)\left( {\frac{1}{{{x^5}}}} \right) + \frac{1}{{{x^6}}} \\ = {x^6} + 6{x^4} + 15{x^2} + 20 + \frac{{15}}{{{x^2}}} + \frac{6}{{{x^4}}} + \frac{1}{{{x^6}}} \\ \end{gathered}\]

Ans. 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, $96 = 100 - 4$ 

\[\begin{gathered} {\left( {96} \right)^3} = {\left( {100 - 4} \right)^3} \\ = {}^3{C_0}{\left( {100} \right)^3} - {}^3{C_1}{\left( {100} \right)^2}\left( 4 \right) + {}^3{C_2}\left( {100} \right){\left( 4 \right)^2} - {}^3{C_3}{\left( 4 \right)^3} \\ = 1000000 - 3\left( {10000} \right)\left( 4 \right) + 3\left( {100} \right)\left( {16} \right) - 64 \\ = 1000000 - 120000 + 4800 - 64 \\ = 884736 \\ \end{gathered}\]

Ans. 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, $102 = 100 + 2$ 

\[\begin{gathered} {\left( {102} \right)^5} = {\left( {100 + 2} \right)^5} \\ = {}^5{C_0}{\left( {100} \right)^5} + {}^5{C_1}{\left( {100} \right)^4}\left( 2 \right) + {}^5{C_2}{\left( {100} \right)^3}{\left( 2 \right)^2} + {}^5{C_3}{\left( {100} \right)^2}{\left( 2 \right)^3} + {}^5{C_4}\left( {100} \right){\left( 2 \right)^4} \\ + {}^5{C_5}{\left( 2 \right)^5} \\ = 10000000000 + 5\left( {100000000} \right)\left( 2 \right) + 10\left( {1000000} \right)\left( 4 \right) + 10\left( {10000} \right)\left( 8 \right) \\ + 5\left( {100} \right)\left( {16} \right) + 32 \\ = 10000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32 \\ = 11040808032 \\ \end{gathered} \]

Ans. 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, $101 = 100 + 1$ 

\[\begin{gathered} {\left( {101} \right)^4} = {\left( {100 + 1} \right)^4} \\ = {}^4{C_0}{\left( {100} \right)^4} + {}^4{C_1}{\left( {100} \right)^3}\left( 1 \right) + {}^4{C_2}{\left( {100} \right)^2}{\left( 1 \right)^2} + {}^4{C_3}\left( {100} \right){\left( 1 \right)^3} + {}^4{C_4}{\left( 1 \right)^4} \\ = 100000000 + 4\left( {1000000} \right) + 6\left( {10000} \right) + 4\left( {100} \right) + 1 \\ = 100000000 + 4000000 + 60000 + 400 + 1 \\ = 104060401 \\ \end{gathered} \]

Ans. 99 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, $99 = 100 - 1$ 

$\begin{gathered} {\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} \\ = 10000000000 - 5\left( {100000000} \right) - 10\left( {1000000} \right) - 10\left( {10000} \right) + 5\left( {100} \right) - 1 \\ = 10000000000 - 500000000 - 10000000 - 100000 + 500 - 1 \\ = 9509900499 \\ \end{gathered} $

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

\[\begin{gathered} {\left( {1.1} \right)^{10000}} = {\left( {1 + 0.1} \right)^{10000}} \\ = {}^{10000}{C_0} + {}^{10000}{C_1}\left( {1.1} \right) + {\text{Other positive terms}} \\ = 1 + 10000 \times 1.1 + {\text{Other positive terms}} \\ = 1 + 11000 + {\text{Other positive terms}} \\ > 1000 \\ \end{gathered}\] Hence, \[{\left( {1.1} \right)^{10000}} > 1000\]