A) \[\frac{1}{n}-\frac{1}{2{{n}^{2}}}+\frac{1}{3{{n}^{3}}}-\frac{1}{4{{n}^{4}}}+...\]
B) \[\frac{1}{n}+\frac{1}{2{{n}^{2}}}+\frac{1}{3{{n}^{3}}}+\frac{1}{4{{n}^{4}}}+...\]
C) \[\frac{1}{n}+\frac{1}{{{2}^{2}}}.\frac{1}{{{n}^{2}}}+\frac{1}{{{2}^{3}}}.\frac{1}{{{n}^{3}}}+...\]
D) None of the above
Correct Answer: A
Solution :
Given,\[\frac{1}{(n+1)}\frac{1}{2{{(n+1)}^{2}}}+\frac{1}{3{{(n+1)}^{3}}}+...\] \[=-\left\{ -\frac{1}{(n+1)}-\frac{1}{2{{(n+1)}^{2}}}-\frac{1}{3{{(n+1)}^{3}}}-...\infty \right\}\] \[=-\log \left\{ 1-\frac{1}{(n+1)} \right\}\] \[=-\log \left\{ \frac{n}{n+1} \right\}\] \[=\log \left( \frac{n+1}{n} \right)=\log \left( 1+\frac{1}{n} \right)\] \[=\frac{1}{n}-\frac{1}{2{{n}^{2}}}+\frac{1}{3{{n}^{3}}}-\frac{1}{4{{n}^{4}}}+...\]
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