NCERT Class 11 Maths Binomial Theorem

NCERT Class 11 Maths Binomial Theorem

Chapter 8 Ex.8.1 Question 1

Expand the expression \({\left(1-2x \right)^5}\)

Solution

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

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

Chapter 8 Ex.8.1 Question 2

Expand the expression \(({\frac{2}{x} - \frac{x}{2})^5}\)

Solution

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

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

Chapter 8 Ex.8.1 Question 3

Expand the expression \({\left( {2}x - {3} \right)^6}\)

Solution

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

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

Chapter 8 Ex.8.1 Question 4

Expand the expression \(\left( \frac{x}{3} + \frac{1}{x} \right)^5\)

Solution

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

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

Chapter 8 Ex.8.1 Question 5

Expand the expression \(\left( {x + \frac{1}{x}} \right)^6\)

Solution

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

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

Chapter 8 Ex.8.1 Question 6

Using Binomial Theorem, evaluate \({\left( 96 \right)^3}\)

Solution

\(96\) can be expressed as the difference of two numbers whose powers are easier to calculate.

Hence, \(96 = 100 - 4\)

Therefore,

\[\begin{align}{\left( {96} \right)^3} &= {\left( {100 - 4} \right)^3}\\&= {}^3{C_0}{\left( {100} \right)^3} - {}^3{C_1}{\left( {100} \right)^2}\left( 4 \right) + {}^3{C_2}\left( {100} \right){\left( 4 \right)^2} - {}^3{C_3}{\left( 4 \right)^3}\\&= {\left( {100} \right)^3} - 3\left( {10000} \right)\left( 4 \right) + 3\left( {100} \right)\left( {16} \right) - \left( {64} \right)\\&= 1000000 - 120000 + 4800 - 64\\&= 1004800 - 120064\\&= 884736\end{align}\]

Chapter 8 Ex.8.1 Question 7

Using Binomial Theorem, evaluate \({\left(102 \right)^5}\)

Solution

\(102\) can be expressed as the sum of two numbers whose powers are easier to calculate.

Hence, \(102 = 100 + 2\)

Therefore;

\[\begin{align}{\left( {102} \right)^5} &= {\left( {100 + 2} \right)^5}\\&= {}^5{C_0}{\left( {100} \right)^5} + {}^5{C_1}{\left( {100} \right)^4}\left( 2 \right) + {}^5{C_2}{\left( {100} \right)^3}{\left( 2 \right)^2}\\&\quad + {}^5{C_3}{\left( {100} \right)^2}{\left( 2 \right)^3} + {}^5{C_4}\left( {100} \right){\left( 2 \right)^4} + {}^5{C_5}\left( 2 \right)^5\\&= 10000000000 + 1000000000 +  40000000\\&\quad+800000 + 8000+ 32\\&= 11040808032\end{align}\]

Chapter 8 Ex.8.1 Question 8

Using Binomial Theorem, evaluate \(\left(101\right)^4\)

Solution

\(101\) can be expressed as the sum of two numbers whose powers are easier to calculate.

Hence \(101 = 100 + 1\)

Therefore,

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

Chapter 8 Ex.8.1 Question 9

Using Binomial Theorem, evaluate \({\left(99\right)^5}\)

Solution

\(99\) can be expressed as the difference of two numbers whose powers are easier to calculate.

Hence, \(99 = 100 - 1\)

Therefore,

\[\begin{align}{\left( {99} \right)^5} &= {\left( {100 - 1} \right)^5}\\&= {}^5{C_0}{\left( {100} \right)^5} - {}^5{C_1}{\left( {100} \right)^4}\left( 1 \right) + {}^5{C_2}{\left( {100} \right)^3}{\left( 1 \right)^2} \\&\quad- {}^5{C_3}{\left( {100} \right)^2}{\left( 1 \right)^3} + {}^5{C_4}\left( {100} \right){\left( 1 \right)^4} - {}^5{C_5}{\left( 1 \right)^5}\\&= {\left( {100} \right)^5} - 5{\left( {100} \right)^4} + 10{\left( {100} \right)^3} - 10{\left( {100} \right)^2} + 5(100) - 1\\&=10000000000 - 500000000 + 10000000 - 100000 +500 -1\\&= 10010000500 - 500100001\\&= 9509900499\end{align}\]

Chapter 8 Ex.8.1 Question 10

Using Binomial Theorem, indicate which number is larger \(\left(1.1 \right)^{10000}\) or \(1000.\)

Solution

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

\[\begin{align}{\left( {1.1} \right)^{10000}} &= {\left( {1 + 0.1} \right)^{10000}}\\&= {}^{10000}{C_0} + {}^{10000}{C_1}\left( {0.1} \right) + other\;positive\;terms\\&= {\rm{1}} + {\rm{1}}0000 \times \left( {0.{\rm{1}}} \right) +other\;positive\;terms\\&= 1 + 1000 + other\;positive\;terms\\&= 1001 + other\;positive\;terms\\&> 1000\end{align}\]

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

Chapter 8 Ex.8.1 Question 11

Find \({\left(a + b\right)^4} - {\left(a - b \right)^4}.\) Hence, evaluate \({\left(\sqrt 3 + \sqrt 2\right)^4} - {\left(\sqrt 3 - \sqrt 2 \right)^4}.\)

Solution

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

\[\begin{align}{\left( {a + b} \right)^4} &= {}^4{C_0}{a^4} + {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} + {}^4{C_3}a{b^3} + {}^4{C_4}{b^4}\\{\left( {a - b} \right)^4} &= {}^4{C_0}{a^4} - {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} - {}^4{C_3}a{b^3} + {}^4{C_4}{b^4}\end{align}\]

Therefore,

\[\begin{align}{\left( {a + b} \right)^4} - {\left( {a - b} \right)^4}& = \left[ \begin{array}{l}\left( {{}^4{C_0}{a^4} + {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} + {}^4{C_3}a{b^3} + {}^4{C_4}{b^4}} \right)\\- \left( {{}^4{C_0}{a^4} - {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} - {}^4{C_3}a{b^3} + {}^4{C_4}{b^4}} \right)\end{array} \right]\\&= 2\left( {{}^4{C_1}{a^3}b + {}^4{C_3}a{b^3}} \right)\\&= 2\left( {4{a^3}b + 4a{b^3}} \right)\\&= 8ab\left( {{a^2} + {b^2}} \right)\end{align}\]

Putting \(a = \sqrt 3 \) and \(b = \sqrt 2 \),

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

Chapter 8 Ex.8.1 Question 12

Find \({\left(x + 1\right)^6} + {\left(x - 1\right)^6}.\) Hence, evaluate \({\left(\sqrt 2 + 1 \right)^6} + {\left( \sqrt 2 - 1 \right)^6}.\)

Solution

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

\[\begin{align}{\left(x + 1 \right)^6} &= {}^6{C_0}{x^6} + {}^6{C_1}{x^5} + {}^6{C_2}{x^4} + {}^6{C_3}{x^3}\\&\quad + {}^6{C_4}{x^2} + {}^6{C_5}x + {}^6{C_6}\\{\left(x - 1 \right)^6} &= {}^6{C_0}{x^6} - {}^6{C_1}{x^5} + {}^6{C_2}{x^4} - {}^6{C_3}{x^3}\\&\quad + {}^6{C_4}{x^2} - {}^6{C_5}x + {}^6{C_6}\end{align}\]

Therefore,

\[\begin{align}{\left(x + 1\right)^6} + {\left(x - 1\right)^6} &= \left[ \begin{array}{l}\left( {{}^6{C_0}{x^6} + {}^6{C_1}{x^5} + {}^6{C_2}{x^4} + {}^6{C_3}{x^3} + {}^6{C_4}{x^2} + {}^6{C_5}x + {}^6{C_6}} \right)\\ + \left( {{}^6{C_0}{x^6} - {}^6{C_1}{x^5} + {}^6{C_2}{x^4} - {}^6{C_3}{x^3} + {}^6{C_4}{x^2} - {}^6{C_5}x + {}^6{C_6}} \right)\end{array} \right]\\&= 2\left[ {{}^6{C_0}{x^6} + {}^6{C_2}{x^4} + {}^6{C_4}{x^2} + {}^6{C_6}} \right]\\&= 2\left[ {{x^6} + 15{x^4} + 15{x^2} + 1} \right]\end{align}\]

Putting \(x = \sqrt 2 \)

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

Chapter 8 Ex.8.1 Question 13

Show that \(9^{n + 1} - 8n - 9\) is divisible by \(64,\) whenever \(n\) is a positive integer.

Solution

To show that \({9^{n + 1}} - 8n - 9\) is divisible by \(64,\) we need to prove that \({9^{n + 1}} - 8n - 9 = 64k\) 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

\[{\left(1 + 8 \right)^{n + 1}} = {}^{n + 1}{C_0} + {}^{n + 1}{C_1}\left( 8 \right) + {}^{n + 1}{C_2}{\left( 8 \right)^2} + \ldots + {}^{n + 1}{C_{n + 1}}{\left( 8 \right)^{n + 1}}\]

Hence,

\[\begin{align}{9^{n + 1}}& = 1 + \left( {n + 1} \right)\left( 8 \right) + {\left( 8 \right)^2}\left[ {{}^{n + 1}{C_2} + {}^{n + 1}{C_3}\left( 8 \right) + ..... + {}^{n + 1}{C_{n + 1}}{{\left( 8 \right)}^{n + 1}}} \right]\\{9^{n + 1}}& = 9 + 8n + 64\left[ {{}^{n + 1}{C_2} + {}^{n + 1}{C_3}\left( 8 \right) + ..... + {}^{n + 1}{C_{n + 1}}{{\left( 8 \right)}^{n + 1}}} \right]\end{align}\]

Therefore, \(9^{n + 1} - 8n - 9 = 64k\)

Where \(k = \left[ {{}^{n + 1}{C_2} + {}^{n + 1}{C_3}\left( 8 \right) + ..... + {}^{n + 1}{C_{n + 1}}{{\left( 8 \right)}^{n + 1}}} \right]\) is a natural number.

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

Chapter 8 Ex.8.1 Question 14

Prove that \(\sum\limits_{r = 0}^n {{3^r}{}^n{C_r} = {4^n}} \)

Solution

By Binomial theorem,

\[\sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r} = {{\left( {a + b} \right)}^n}}\]

By putting \(b = 3\) and \(a = 1\), we obtain

\[\begin{align}\sum\limits_{r = 0}^n {{}^n{C_r}{\left( 1 \right)}^{n - r}}{3^r} &= {{(1 + 3)}^n} \\\sum\limits_{r = 0}^n {3^r}{}^n{C_r} &= {4^n} \end{align}\]

Hence proved.

Binomial Theorem | NCERT Solutions
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