question_answer

\[\frac{1}{2}{{x}^{2}}+\frac{2}{3}{{x}^{3}}+\frac{3}{4}{{x}^{4}}+\frac{4}{5}{{x}^{5}}+.....\]is

A)  \[\frac{-x}{1+x}+log\left( 1+x \right)\]  

B)  \[\frac{x}{1+x}+log\left( 1+x \right)\]

C)  \[\frac{x}{1-x}+log\left( 1+x \right)\]     

D)  \[\frac{1+x}{1-x}+log\left( 1+x \right)\]

Correct Answer: C

Solution :

[c] \[\because \frac{1}{2}{{x}^{2}}+\frac{2}{3}{{x}^{3}}+\frac{3}{4}{{x}^{4}}+\frac{4}{5}{{x}^{5}}+......\infty \] \[{{t}_{n}}\frac{n}{n+1}.{{x}^{n+1}}=\frac{n+1-1}{n+1}\times {{x}^{n+1}}=\left( 1-\frac{1}{n+1} \right){{x}^{n+1}}={{x}^{n+1}}-\frac{{{x}^{n+1}}}{n+1}\] \[{{S}_{n}}=\sum\limits_{n=1}^{\infty }{{{t}_{n}}=({{x}^{2}}+{{x}^{3}}+{{x}^{4}}+.....\infty )}\] \[-\left( \frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+\frac{{{x}^{4}}}{4}+.....\infty  \right)\] \[=(x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}+..\infty )-\left( x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+\frac{{{x}^{4}}}{4}+...\infty  \right)\]