Question

Find the sum of the n terms of the sequence
$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+.............n^2$

Answer 1

$\begin{array}{c}\frac{n}{1+n^2+n^4}\\ =\frac{n}{{(n^2+1)}^2-n^2}(since,n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1{)}^2-n^2)\\ =\frac{n}{(n^2-n+1)(n^2+1+n)}\\ =\frac{1}{2}\left[\frac{2n}{(n^2-n+1)(n^2+1+n)}\right]\\ =\frac{1}{2}\left[\frac{-1}{n^2+n+1}+\frac{1}{n^2-n+1}\right]\\ \sum \frac{n}{n^4+n^2+1}={\sum }_{n=1}^n\frac{1}{2}\left[\left(\frac{-1}{n^2+n+1}\right)+\left(\frac{1}{n^2-n+1}\right)\right]\\ =\frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+....\frac{-1}{n^2+n+1}\right]\\ =\frac{1}{2}\left[1-\frac{1}{n^2+n+1}\right]\\ =\frac{1}{2}\left(\frac{n^2+n}{n^2+n+1}\right)\end{array}$