• # question_answer The sum of the series $\frac{1}{1+{{1}^{2}}+{{1}^{4}}}+\frac{2}{1+{{2}^{2}}+{{2}^{4}}}+\frac{3}{1+{{3}^{2}}+{{3}^{4}}}+.........$ to $n$ terms is A) $\frac{n({{n}^{2}}+1)}{{{n}^{2}}+n+1}$ B) $\frac{n(n+1)}{2({{n}^{2}}+n+1)}$ C) $\frac{n({{n}^{2}}-1)}{2({{n}^{2}}+n+1)}$ D) None of these

Correct Answer: B

Solution :

Let ${{T}_{n}}$ be the ${{n}^{th}}$ term of the series $\frac{1}{1+{{1}^{2}}+{{1}^{4}}}+\frac{2}{1+{{2}^{2}}+{{2}^{4}}}+\frac{3}{1+{{3}^{2}}+{{3}^{4}}}+......$ Then ${{T}_{n}}=\frac{n}{1+{{n}^{2}}+{{n}^{4}}}=\frac{n}{{{(1+{{n}^{2}})}^{2}}-{{n}^{2}}}$            $=\frac{n}{({{n}^{2}}+n+1)({{n}^{2}}-n+1)}$           $=\frac{1}{2}\left[ \frac{1}{{{n}^{2}}-n+1}-\frac{1}{{{n}^{2}}+n+1} \right]$           $=\frac{1}{2}\left[ \frac{1}{1+(n-1)n}-\frac{1}{1+n(n+1)} \right]$ Now  $\sum\limits_{r=1}^{n}{{{T}_{r}}}=\frac{1}{2}\left[ \frac{1}{1}-\frac{1}{1+1.2} \right]+\frac{1}{2}\left[ \frac{1}{1+1.2}-\frac{1}{1+2.3} \right]$ $+\frac{1}{2}\left[ \frac{1}{1+2.3}-\frac{1}{1+3.4} \right]+.......+\frac{1}{2}\left[ \frac{1}{1+(n-1)n}-\frac{1}{1+n(n+1)} \right]$ $=\frac{1}{2}\left[ 1-\frac{1}{1+n(n+1)} \right]=\frac{n(n+1)}{2({{n}^{2}}+n+1)}$. Trick: Checking for$c$. ${{S}_{1}}=\frac{1}{3}$ and ${{S}_{2}}=\frac{3}{7}$ which are given by (b).

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