Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is the second of our many quizzes on Binomial theorem. Click on this to access more such competitive quizzes on Algebra.

Binomial theorem Quiz 2

mathematics

  What will be the binomial expansion of  ${{(b+2)}^{4}}$.

 By applying a binomial formula, we can get the binomial expansion of the given expression.

  $\text{We know,}\\ {{(a+b)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}b{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...{{+}^{n}}{{C}_{n}}{{b}^{n}}\\ \text{Therefore, by expanding } {{(b+2)}^{4}},\\{{(b+2)}^{4}}{{=}^{4}}{{C}_{0}}{{b}^{4}}{{+}^{4}}{{C}_{1}}{{b}^{4-1}}(2){{+}^{4}}{{C}_{2}}{{b}^{4-2}}{{(2)}^{2}}{{+}^{4}}{{C}_{3}}{{b}^{4-3}}{{(2)}^{3}}{{+}^{4}}{{C}_{4}}{{b}^{4-4}}{{(2)}^{4}} \\ \Rightarrow {{(b+2)}^{4}}=(1){{b}^{4}}+(4){{b}^{3}}(2)+(6){{b}^{2}}(4)+(4){{b}^{1}}(8)+(1){{b}^{0}}(16) \\ \therefore {{(b+2)}^{4}}={{b}^{4}}+8{{b}^{3}}+24{{b}^{2}}+32b+16 $

 What is the coefficient of  ${{a}^{3}}$  in the expansion of  ${{(a+4)}^{5}}$.

 By applying pascal's triangle, we get the coefficient of ${{a}^{3}}$ using the above expansion.

  $\text{We have,} \\{{\left( a+4 \right)}^{5}} \\ \text{To find } {{a}^{3}},\\ \text{we have}\\ ^{5}{{C}_{3}}{{\left( a \right)}^{3}}{{\left( 4 \right)}^{2}}=\dfrac{5!}{3!}{{a}^{3}}(16)=160{{a}^{3}}\\ \text{Therefore, the coefficient is 160.}$

 What is the difference between the coefficient of  ${{a}^{6}}$  and  ${{a}^{4}}$ in the expansion of  ${{(4a+b)}^{6}}$.

 By applying pascal's triangle, we get the coefficient of ${{a}^{6}},{{a}^{4}}$ using the above expansion. Then find the difference.

$\text{We have},\, {{(4a+b)}^{6}}\\ \text{To find the coefficient of } a^6, \text{ we have}\\ {{\Rightarrow }^{6}}{{C}_{6}}{{(4a)}^{6}}.b \\ \Rightarrow 4096{{a}^{6}}.b \\ \text{Coefficient of } {{a}^{6}}=4096\\ \text{We have, } \\{{(4a+b)}^{6}}\\ \text{To find the coefficient of } a^4, \text{ we have } {{\Rightarrow }^{6}}{{C}_{4}}{{(4a)}^{4}}.b \\  \Rightarrow 3840{{a}^{4}}.b \\ \text{ Coefficient of } {{a}^{4}}=3840 \\ \text{Therefore, the difference is 4096-3840=256.}$

 What is the expansion of  ${{\left( 2a(1-b) \right)}^{3}}$.

 For the expansion of the given expression, we have to use a binomial formula. We can either split the given expression or open the bracket and apply the binomial expansion formula.

 ${{\left( 2a(1-b) \right)}^{3}}={{\left( 2a \right)}^{3}}{{\left( 1+(-b) \right)}^{3}} \\ \text{Expanding the given expression with binomial formula, we get}\\ ={{\left( 2a \right)}^{3}}\left[ \dfrac{3!}{(3-0)!0!}.{{(-b)}^{3-0}}{{(1)}^{0}}+\dfrac{3!}{(3-1)!1!}.{{(-b)}^{3-1}}{{(1)}^{1}}+\dfrac{3!}{(3-2)!2!}.{{(-b)}^{3-2}}{{(1)}^{2}}+\dfrac{3!}{(3-3)!3!}.{{(-b)}^{3-3}}{{(1)}^{3}} \right] \\  =8{{a}^{3}}\left[ 1-3b+3{{b}^{2}}-{{b}^{3}} \right] \\ =-8{{a}^{3}}{{b}^{3}}+24{{a}^{3}}{{b}^{2}}-24{{a}^{3}}b+8{{a}^{3}} $

 Find the value of  ${{(2+a)}^{4}}+{{(3-a)}^{4}}$.

 Expand the expression separately using binomial expansion then add them to get the required answer.

 $\text{We know},\\ {{(a+b)}^{4}}={{a}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+{{b}^{4}} \\ \therefore {{(2+a)}^{4}}={{2}^{4}}+4{{(2)}^{3}}a+6{{(2)}^{2}}{{a}^{2}}+4(2){{a}^{3}}+{{a}^{4}} \\ \Rightarrow {{(2+a)}^{4}}=16+32a+24{{a}^{2}}+8{{a}^{3}}+{{a}^{4}} \\ \text{Similarly,} \\ {{(a-b)}^{4}}={{a}^{4}}-4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}-4a{{b}^{3}}+{{b}^{4}} \\ \therefore {{(3-a)}^{4}}={{3}^{4}}-4{{(3)}^{3}}a+6{{(3)}^{2}}{{a}^{2}}-4(3){{a}^{3}}+{{a}^{4}} \\ \Rightarrow {{(3-a)}^{4}}=81-108a+54{{a}^{2}}-12{{a}^{3}}+{{a}^{4}} \\ \text{Therefore, the required value is} \\ {{(2+a)}^{4}}+{{(3-a)}^{4}}=16+32a+24{{a}^{2}}+8{{a}^{3}}+{{a}^{4}}+81-108a+54{{a}^{2}}-12{{a}^{3}}+{{a}^{4}} \\ \Rightarrow {{(2+a)}^{4}}+{{(3-a)}^{4}}=2{{a}^{4}}-4{{a}^{3}}+78{{a}^{2}}-76a+97 \\ \therefore {{(2+a)}^{4}}+{{(3-a)}^{4}}=2\left( {{a}^{4}}-2{{a}^{3}}+39{{a}^{2}}-38a+\dfrac{97}{2} \right) $