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Looking for Class 11 Maths Exercise 8.1 solutions? You’re in the right place! This section offers detailed and step-by-step solutions for all questions from Exercise 8.1 of Chapter 8 – Binomial Theorem. These solutions are prepared according to the latest NCERT curriculum and help you understand the expansion of binomial expressions, binomial coefficients, and general terms of the expansion. Whether you’re revising from the Class 11 Ch 8 Exercise 8.1 solutions, practicing with the NCERT Exemplar Class 11 Maths, or strengthening your understanding of the Binomial Theorem Class 11, these explanations are perfect for building a solid foundation. View or download the complete NCERT Solutions for Class 11 Maths Chapter 8 and master the binomial theorem with confidence!
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EXERCISE 8.1
1. Expand each of the expressions in \( (1-2 x)^{5} \)
Answer
By using binominal theorem, the expression \( (1-2 x)^{5} \) can be expanded as \( (1-2 {x})^{5} \)
\(= { }^{5} {C}_{0}(1)^{5}-{ }^{5} {C}_{1}(1)^{4}(2 {x})+{ }^{5} {C}_{2}(1)^{3}(2 {x})^{2}-{ }^{5} {C}_{3}(1)^{2}(2 {x})^{3}\) \(+{ }^{5} {C}_{4}(1)^{1}(2 {x})^{4}- { }^{5} {C}_{5}(2 {x})^{5}\)
\(= 1-5(2 {x})+10(4 {x})^{2}-10(8 {x})^{3}+5(16 {x})^{4}-(32 {x})^{5}\)
\(= 1-10 {x}+40 {x}^{2}-80 {x}^{3}+80 {x}^{4}-32 {x}^{5}\)
2. Expand each of the expressions in \( \left(\frac{2}{x}-\frac{x}{2}\right)^{5} \)
Answer
By using binomial theorem, the expression \( \left(\frac{2}{x}-\frac{x}{2}\right)^{5} \) can be expanded as
\(\left\{\frac{2}{x}-\frac{x}{2}\right\}^{5}= {}^{5} {C}_{0}\left(\frac{2}{x}\right)^{5}-{}^{5} {C} _{1}\left(\frac{2}{x}\right)^{4}\left(\frac{x}{2}\right)+{}^{5} {C}_{2}\left(\frac{2}{x}\right)^{3}\left(\frac{x}{2}\right)^{2}\) \(-{}^{5} {C} _{3}\left(\frac{2}{x}\right)^{2}\left(\frac{x}{2}\right)^{3}+{}^{5} {C} _{4}\left(\frac{2}{x}\right)\left(\frac{x}{2}\right)^{4}-{}^{5} C_{5}\left(\frac{x}{2}\right)^{5}\)
\(=\frac{32}{x^{5}}-5\left(\frac{16}{x^{4}}\right)\left(\frac{x}{2}\right)+10\left(\frac{8}{x^{3}}\right)\left(\frac{x^{2}}{4}\right)-10\left(\frac{4}{x^{2}}\right)\left(\frac{x^{2}}{8}\right)+5\left(\frac{2}{x}\right)\left(\frac{x^{4}}{16}\right)-\frac{x^{5}}{32}\)
\(=\frac{32}{x^{5}}-\frac{40}{x^{3}}+\frac{20}{x}-5 x+\frac{5}{8} x^{2}-\frac{x^{5}}{32}\)
3. Expand each of the expressions in \( (2 x-3)^{6} \)
Answer
By using binomial theorem, the expression \( (2 {x}-3)^{6} \) can be expended as
\((2 {x}-3)^{6}={ }^{6} {C}_{0}(2 {x})^{6}-{ }^{6} {C}_{1}(2 {x})^{5}(3)+{ }^{6} {C}_{2}(2 {x})^{4}(3)^{2}\) \(-{ }^{6} {C}_{3}(2 {x})^{3}(3)^{3}+{ }^{6} {C}_{4}(2 {x})^{2}(3)^{4}-{ }^{6} {C}_{5}(2 {x})(3)^{5}+{ }^{6} {C}_{6}(3)^{6}\)
\(=64 {x}^{6}-6\left(32 {x}^{5}\right)(3)+15\left(16 {x}^{4}\right)(9)-20\left(8 {x}^{3}\right)(27)\) \(+15\left(4 {x}^{2}\right)(81)-6(2 {x})(243)+729\)
\(=64 {x}^{6}-576 {x}^{5}+2160 {x}^{4}-4320 {x}^{3}+4860 {x}^{2}-2916 {x}+729\)
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4. Expand each of the expressions in \( \left\{\frac{x}{3}+\frac{1}{X}\right\}^{5} \)
Answer
By using binomial theorem, the expression \( \left\{\frac{X}{3}+\frac{1}{X}\right\}^{5} \) can be expanded as
\(\left\{\frac{X}{3}+\frac{1}{X}\right\}^{5}={}^{5} {C}_{0}\left(\frac{X}{3}\right)^{5}+{}^{5} {C}_{1}\left(\frac{X}{3}\right)^{4}\left(\frac{1}{X}\right)+{}^{5} {C}_{2}\left(\frac{X}{3}\right)^{3}\left(\frac{1}{X}\right)^{2}\) \(+{}^{5} {C}_{3}\left(\frac{X}{3}\right)^{2}\left(\frac{1}{X}\right)^{3}+{}^{5} {C}_{4}\left(\left(\frac{X}{3}\right)\right)\left(\frac{1}{X}\right)^{4}+{}^{5} {C}_{5}\left(\frac{1}{X}\right)^{5}\)
\(=\frac{X^{5}}{243}+5\left(\frac{X^{4}}{81}\right)\left(\frac{1}{X}\right)+10\left(\frac{X^{3}}{27}\right)\left(\frac{1}{X^{2}}\right)+10\left(\frac{X^{2}}{9}\right)\left(\frac{1}{X^{3}}\right)\) \(+5\left(\frac{X}{3}\right)\left(\frac{1}{X^{4}}\right)+\frac{1}{X^{5}}\)
\(=\frac{X^{5}}{243}+\frac{5 X^{3}}{81}+\frac{10 X}{9 X}+\frac{5}{3 X^{3}}+\frac{1}{X^{5}}\)
5. Expand each of the expressions in \( \left\{x+\frac{1}{x}\right\}^{6} \)
Answer
By using binomial theorem, the expression \( \left\{x+\frac{1}{x}\right\}^{6} \) can be expanded as
\(\left\{x+\frac{1}{x}\right\}^{6}={ }^{5} {C}_{0}({x})^{6}+{ }^{5} {C}_{1}({x})^{5}\left(\frac{1}{x}\right)+{ }^{6} {C}_{2}({x})^{4}\left(\frac{1}{x}\right)^{2}\) \(+{ }^{6} {C}_{3}({x})^{3}\left(\frac{1}{x}\right)^{3}+{ }^{6} {C}_{4}({x})^{2}\left(\frac{1}{x}\right)^{4}+{ }^{6} {C}_{5}({x})\left(\frac{1}{x}\right)^{4}+{ }^{6} {C}^{6}\left(\frac{1}{x}\right)^{4}\)
\(= x^{6}+6(x)^{5}\left(\frac{1}{x}\right)+15(x)^{4}\left(\frac{1}{x^{2}}\right)+20(x)^{3}\left(\frac{1}{x^{3}}\right)+15(x)^{2}\left(\frac{1}{x^{4}}\right)\) \(+6({x})\left(\frac{1}{x^{5}}\right)+\frac{1}{x^{6}}\)
\(= x^{6}+6 x^{4}+15 x^{2}+20+\frac{15}{x^{2}}+\frac{6}{x^{4}}+\frac{1}{x^{6}}\)
6. Using binomial theorem, evaluate each of the following: \( (96)^{3} \)
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, \( 96=100-4 \)
\((96)^{3}=(100-4)^{3}\)
\(={ }^{3} {C}_{0}(100)^{3}-{ }^{3} {C}_{1}(100)^{2}(4)+{ }^{3} {C}_{2}(100)(4)^{2}-{ }^{3} {C}_{2}(4)^{3}\)
\(=(100)^{3}-3(100)^{2}(4)+3(100)(4)^{2}-(4)^{3}\)
\(=1000000-120000+4800-64\)
\(=884736\)
7. Using binomial theorem, evaluate each of the following: \((102) ^{5} \)
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, \( 102=100+2 \)
\((102)^{5}=(100+2)^{5}\)
\(={ }^{5} {C}_{0}(100)^{5}+{ }^{5} {C}_{1}(100)^{4}(2)+{ }^{5} {C}_{2}(100)^{3}(2)^{2}+{ }^{5} {C}_{3}(100)^{4}(2)^{3}\) \(+{ }^{5} {C}_{4}(100)(2)^{4}+{ }^{5} {C}_{5}(2)^{5}\)
\(=1000000000+100000000+40000000+80000+800032\)
\(=11040808032\)
8. Using binomial theorem, evaluate each of the following: \( (101) ^{4} \)
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, \( 101=100+1 \)
\((101)^{4}=(100+1)^{4}\)
\(={ }^{4} {C}_{0}(100)^{4}+{ }^{4} {C}_{1}(100)^{3}(1)+{ }^{4} {C}_{2}(100)^{2}(1)^{2}+{ }^{4} {C}_{3}(100)(1)^{3}+{ }^{4} {C}_{4}(1)^{4}\)
\(=(100)^{4}+4(100)^{3}+6(100)^{2}+4(100)+(1)^{4}\)
\(=10000000+4000000+60000+400+1\)
\(=104060401\)
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9. Using binomial theorem, evaluate each of the following: \((99) ^{5} \)
Answer
99 can be written as the sum of difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.
It can be written that, \( 99=100-1 \)
\((99)^{5}=(100-1)^{5}\)
\(={ }^{5} {C}_{0}(100)^{5}-{ }^{5} {C}_{1}(100)^{4}(1)+{ }^{5} {C}_{2}(100)^{3}(1)^{2}-{ }^{5} {C}_{3}(100)^{2}(1)^{3}\) \(+{ }^{5} {C}_{4}(100)(1)^{4}-{ }^{5} {C}_{5}(1)^{5}\)
\(=(100)^{5}-5(100)^{4}+10(100)^{3}-10(100)^{2}+5(100)-1\)
\(=10000000000-500000000+1000000-100000+500-1\)
\(=10010000500-500100001\)
\(=9509900499\)
10. Using Binomial Theorem, indicate which number is larger \( (1.1)^{10000} \) or \(1000 \).
Answer
By splitting 1.1 and then applying binomial theorem, the first few term of \( (1.1)^{10000} \) or \(1000 \).
as
\((1.1)^{10000}=(1+0.1) 1000\)
\(={ }^{10000} {C}_{0}+{ }^{10000} {C}_{2}(1.1)+\text { other positive terms }\)
\(=1+10000 \times 1.1+\text { other positive terms }\)
\(=1+10000+\text { other positive terms } > 1000\)
Hence, \((1.1) 10000 > 1000 \).
11. Find \( (a+b)^{4}-(a-b)^{4} \). Hence, evaluate \( (\sqrt{3}+\sqrt{2})^{4}- (\sqrt{3}-\sqrt{2})^{4} \)
Answer
Using binomial theorem, the expressions, \( (a+b)^{4}\) and \( (a-b)^{4} ,\) can be expanded as
\(({a}+{b})^{4}={ }^{4} {C}_{0} {a}^{4}+{ }^{4} {C}_{1} {a}^{3} {~b}+{ }^{4} {C}_{2} {a}^{2} {~b}^{2}+{ }^{4} {C}_{3} {ab}^{3}+{ }^{4} {C}_{4} {b}^{4}\)
\(({a}-{b})^{4}={ }^{4} {C}_{0} {a}^{4}-{ }^{4} {C}_{1} {a}^{3} {b}+{ }^{4} {C}_{2} {a}^{2} {b}^{2}-{ }^{4} {C}_{3} {ab}^{3}+{ }^{4} {C}_{4} {b}^{4}\)
\(({a}+{b})^{4}-({a}-{b})^{4}={ }^{4} {C}_{0} {a}^{4}+{ }^{4} {C}_{1} {a}^{3} {b}+{ }^{4} {C}_{2} {a}^{2} {b}^{2}+{ }^{4} {C}_{3} {ab}^{3}+{ }^{4} {C}_{4} {b}^{4}\)
\(-\left[{ }^{4} {C}_{0} {a}^{4}-{ }^{4} {C}_{1} {a}^{3} {b}+{ }^{4} {C}_{2} {a}^{2} {b}^{2}-{ }^{4} {C}_{3} {ab}^{3}+{ }^{4} {C}_{4} {b}^{4}\right]\)
\(=2\left({ }^{4} {C}_{2} {a}^{3} {b}+{ }^{4} {C}_{3} {ab}^{3}\right)=2\left(4 {a}^{3} {b}+4 {ab}^{3}\right)\)
\(=8 {ab}\left({a}^{2}+{b}^{2}\right)\)
By putting \( {a}=\sqrt{3} \) and \( {b}=\sqrt{2} \), we obtain
\((\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2}=8(\sqrt{3})(\sqrt{2})\left\{(\sqrt{3})^{2}+(\sqrt{2})^{2}\right\}\)
\(=8(\sqrt{6})[3+2]=40 \sqrt{6}\)
12. Find \( (x+1)^{6}+(x-1)^{6} \). Hence or otherwise evaluate \( (\sqrt{2}+1)^{6}+ (\sqrt{2}-1)^{6} \)
Answer
Using binomial theorem, the expression, \( (x+1)^{6} \) and \( (x-1)^{6} \), can be expanded as
\(({x}+1)^{6}={ }^{6} {C}_{0} {x}^{6}+{ }^{6} {C}_{1} {x}^{5}+{ }^{6} {C}_{2} {x}^{4}+{ }^{6} {C}_{3} {x}^{3}+{ }^{6} {C}_{4} {x}^{2}+{ }^{6} {C}_{5} {x}+{ }^{6} {C}_{6}\)
\(({x}-1)^{6}={ }^{6} {C}_{0} {x}^{5}-{ }^{6} {C}_{1} {x}^{5}+{ }^{6} {C}_{2} {x}^{4}-{ }^{6} {C}_{3} {x}^{3}-{ }^{6} {C}_{2} {x}^{2}+{ }^{6} {C}_{4} {x}^{2}-{ }^{6} {C}_{5} {x}^{6}+{ }^{6} {C}_{6}\)
\(({x}+1)^{6}+({x}-1)^{6}=2\left[{ }^{6} {C}_{0} {x}^{6}+{ }^{6} {C}_{2} {x}^{4}+{ }^{6} {C}_{4} {x}^{2}+{ }^{6} {C}_{6}\right]\)
\(=2\left[{x}^{6}+15 {x}^{4}+15 {x}^{2}+1\right]\)
By putting \( x=\sqrt{2} \), we obtain
\((\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}=2\left[(\sqrt{2})^{6}+15(\sqrt{2})^{4}+15(\sqrt{2})^{2}+1\right]\)
\(=2(8+15 \times 4+15 \times 2+1)\)
\(=2(8+60+30+1)\)
\(=2(99)=198\)
13. Show that \( 9^{n+1}-8 n-9 \) is divisible by \(64 \), whenever \( n \) is a positive integer.
Answer
In order to show that \( 9 n-1-8 n-9 \) is divisible by \(64 \), it has to be prove that, \( 9 {n}-1-8 {n}-9=64 {k} \)
Where \(k\) is some natural number
By binomial theorem,
\((1+a)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} a+{ }^{n} C_{2} a _{2}+\ldots+{ }^{n} C n \text { an }\)
For \( {a}=8 \) and \( {m}={n}+1 \), we obtain
\((1+8)^{n+1}={ }^{n+1} C_{0}+{ }^{n+1} C_{1}(8)+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}\)
\(=9^{n+1}=9+8 n+64\left\{{ }^{n+1} C_{2}+{ }^{n+1} C_{3} \times 8+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}\right\}\)
\( =9 n+1-8 n-9=64 k, \text{ where }k={ }^{n+1} C_{2}+{ }^{n+1} C_{3} \times 8\) \(+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1} \) is a natural number.
Thus, \( 9 n+1-8 n-9 \) is divisible by \(64 \), whenever \( n \) is a positive integer.
14. Prove that \(\sum_{r=0}^{n} 3^{r} n_{c_{r}}=4^{n}\)
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
\(\sum_{r=0}^{n} n_{c_{r}}(1)^{{n}-{r}}(3)^{{r}}=(1+3)^{{n}}\)
\(=\sum_{r=0}^{n} 3^{r}{ }^{{n}} C r=4^{{n}}\)
Hence, proved
