Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Match the following.
Column I Column II
(i) $\left(1-2x\right)^{5}=$ (p) $\frac{x^{5}}{243} -\frac{5x^{3}}{81}+\frac{10x}{27}-\frac{10}{9x}$
$+\frac{5}{3x^{3}}-\frac{1}{x^{5}}$
(ii) $\left(2x-3\right)^{6}=$ (q) $x^{6}+6x^{4}+15x^{2}+20$
$+\frac{15}{x^{2}}+\frac{6}{x^{4}}+\frac{1}{x^{6}}$
(iii) $\left(\frac{x}{3}-\frac{1}{x}\right)^{5} =$ (r) $1 - 10x + 40x^{2 }- 80x^{3}$
$+ 80x^{4 }- 32x^{5}$
(iv) $\left(x-\frac{1}{x}\right)^{6} =$ (s) $64x^{6} - 576x^{5} + 2160x^{4} - 4320x^{3}$
$+ 4860x^{2} - 2916x + 729$

Binomial Theorem

Solution:

(i) $\left(1-2x\right)^{5} = \,{}^{5}C_{0}\left(1\right)^{5}- \,{}^{5}C_{1}\left(1\right)^{4}\,2x+ \,{}^{5}C_{2}\left(1\right)^{3}\left(2x\right)^{2}$
$- \,{}^{5}C_{3}\left(1\right)^{2}\left(2x\right)^{3}+ \,{}^{5}C_{4}\left(1\right)\left(2x\right)^{4}- \,{}^{5}C_{5}\left(2x\right)^{5}$
$= 1-5\times2x +\frac{5\times 4}{2}\times 4x^{2}-\frac{5\times 4}{2}\times 8x^{3}$
$+5\times 16x^{4}-32x^{5}$
$= 1 - 10x + 40x^{2} - 80x^{3} + 80x^{4} - 32x^{5}$
(ii) $\left(2x - 3\right)^{6} = \,{}^{6}C_{0}\left(2x\right)^{6} - \,{}^{6}C_{1}\left(2x\right)^{5} \times 3 + \,{}^{6}C_{2}\left(2x\right)^{4}\left(3\right)^{2}$
$- \,{}^{6}C_{3}\left(2x\right)^{3}\left(3\right)^{3}+ \,{}^{6}C_{4}\left(2x\right)^{2}\left(3\right)^{4}- \,{}^{6}C_{5}\left(2x\right)3^{5}+ \,{}^{6}C_{6} 3^{6}$
$= 64x^{6 }-6 \times 32 \times 3 \times x^{5} + \frac{6\times 5}{2}\times 16x^{4}\times 9$
$-\frac{6\times 5\times 4\times 8x^{3}\times 27}{6}+\frac{6\times 5}{2}\times 4x^{2}\times 81$
$-6\times 2x\times 243+729$
$= 64x^{6}-576x^{5}+2160x^{4}-4320x^{3}+4860x^{2}-2916x+729$
(iii) $\left(\frac{x}{3}-\frac{1}{x}\right)^{5}=\,{}^{5}C_{0}\left(\frac{x}{3}\right)^{5}-\,{}^{5}C_{1}\left(\frac{x}{3}\right)^{4}+\,{}^{5}C_{2}\left(\frac{x}{3}\right)^{3}\left(\frac{1}{x}\right)^{2}$
$-\,{}^{5}C_{3}\left(\frac{x}{3}\right)^{2}\left(\frac{1}{x}\right)^{3}+\,{}^{5}C_{4}\left(\frac{x}{3}\right) \left(\frac{1}{x}\right)^{4}-\,{}^{5}C_{5}\left(\frac{x}{3}\right)^{5}$
$= \frac{x^{5}}{243}-\frac{5x^{3}}{81}+\frac{10x}{27}-\frac{10}{9x}+\frac{5}{3x^{3}}-\frac{1}{x^{5}}$
(iv) $\left(x +\frac{1}{x}\right)^{6}=\,{}^{6}C_{0}\,x^{6}+\,{}^{6}C_{1}\,x^{5} \left(\frac{1}{x}\right)+\,{}^{6}C_{2}\,x^{4}\left(\frac{1}{x}\right)^{2}$
$+\,{}^{6}C_{3}\,x^{3}\left(\frac{1}{x}\right)^{3}+\,{}^{6}C_{4}\,x^{2}\left(\frac{1}{x}\right)^{4}+\,{}^{6}C_{5}\,\left(x\right) \left(\frac{1}{x}\right)^{5}+\,{}^{6}C_{6} \left(\frac{1}{x}\right)^{6}$
$= x^{6 }+ 6x^{4}+ 15x^{2} + 20+\frac{15}{x^{2}}+\frac{6}{x^{4}}+\frac{1}{x^{6}}$