Question

Sum to $n$: $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}$

Answer 1

let $S_n=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+$---n

$tn=\frac{r}{1+r^2+r^4}$

$tn=\frac{r}{r^4+r^2+1}$

$tn=\frac{r}{(r^2+r+1)(r^2-r+1)}$

$tn=\frac{1}{2}\times \frac{2r}{(r^2+r+1)-(r^2-r+1)}$

$=\frac{1}{2}\times \frac{\left[\right(r^2+r+1)-(r^2-r+1\left)\right]}{(r^2+r+1)(r^2-r+1)}$

$tn=\frac{1}{2}[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}]$

$S_n=\sum _{r=1}^{n}tn$

$=\frac{1}{2}[\frac{1}{1^2-1+1}-\frac{1}{1^2+1+1}+\frac{1}{2^2-2+1}-\frac{1}{2^2+2+1}+...+\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}$

$S_n=\frac{1}{2}[1-\frac{1}{n^2+n+1}]$

$\Rightarrow S_n=\frac{1}{2}\frac{(n^2+n+1-1)}{n^2+n+1}$

$S_n=\frac{n(n+1)}{2(n^2+n+1)}$