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] \[\because \frac{1}{2}{{x}^{2}}+\frac{2}{3}{{x}^{3}}+\frac{3}{4}{{x}^{4}}+\frac{4}{5}{{x}^{5}}+......\infty \] \[{{t}_{n}}\frac{n}{n+1}.{{x}^{n+1}}=\frac{n+1-1}{n+1}\times {{x}^{n+1}}=\left( 1-\frac{1}{n+1} \right){{x}^{n+1}}={{x}^{n+1}}-\frac{{{x}^{n+1}}}{n+1}\] \[{{S}_{n}}=\sum\limits_{n=1}^{\infty }{{{t}_{n}}=({{x}^{2}}+{{x}^{3}}+{{x}^{4}}+.....\infty )}\] \[-\left( \frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+\frac{{{x}^{4}}}{4}+.....\infty \right)\] \[=(x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}+..\infty )-\left( x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+\frac{{{x}^{4}}}{4}+...\infty \right)\]You need to login to perform this action.
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