Question

The sum to $n$ terms of the series:
$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\cdots$ is

• 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

Answer 1

The correct option is C $\frac{n^2+n}{2(n^2+n+1)}$

Let $T_r$ be the $r^{th}$ term of the given series.
Then, $T_r=\frac{r}{1+r^2+r^4},$ $r=1,2,3,\dots ,n$
$=\frac{r}{(r^2+r+1)(r^2-r+1)}$
$=\frac{1}{2}[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}]$
Sum of the series $=\sum _{r=1}^n{T}_r$
$=\frac{1}{2}\left\{\sum _{r=1}^n(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1})\right\}$
$=\frac{1}{2}\{(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{13})+\cdots +(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1})\}$
$=\frac{1}{2}\{1-\frac{1}{n^2+n+1}\}=\frac{n^2+n}{2(n^2+n+1)}$