A) \[\frac{{{n}^{2}}+1}{2({{n}^{2}}+n+1)}\]
B) \[\frac{{{n}^{2}}+n}{({{n}^{2}}+n+1)}\]
C) \[\frac{{{n}^{2}}+n}{2({{n}^{2}}+n+1)}\]
D) None of these
Correct Answer: C
Solution :
Let \[{{T}_{r}}\]be the\[{{r}^{th}}\]term of the given series. Then \[{{T}_{r}}=\frac{r}{1+{{r}^{2}}+{{r}^{4}}}\], \[r=1,\,\,2,\,\,3,...,n\] \[=\frac{r}{({{r}^{2}}+r+1)({{r}^{2}}-r+1)}\] \[=\frac{1}{2}\left[ \frac{1}{{{r}^{2}}-r+1}-\frac{1}{{{r}^{2}}+r+1} \right]\] \[\therefore \]Sum of the series\[=\sum\limits_{r=1}^{n}{{{T}_{r}}}\] \[=\frac{1}{2}\left\{ \sum\limits_{r=1}^{n}{\left( \frac{1}{{{r}^{2}}-r+1}-\frac{1}{{{r}^{2}}+r+1} \right)} \right\}\] \[=\frac{1}{2}\left\{ \left( 1-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{7} \right)+\left( \frac{1}{7}-\frac{1}{13} \right)+...+ \right.\] \[\left. \left( \frac{1}{{{n}^{2}}-n+1}-\frac{1}{{{n}^{2}}+n+1} \right) \right\}\] \[=\frac{1}{2}\left\{ 1-\frac{1}{{{n}^{2}}+n+1} \right\}=\frac{{{n}^{2}}+n}{2({{n}^{2}}+n+1)}\]
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