Find the sum value: $$\sum^{50}_{k=1} \frac{k}{1+k^2+k^4}$$
Attempt: = $$\sum^{50}_{k=1}\frac{k}{1+k^2+k^4+k^2-k^2}= \sum^{50}_{k=1} \frac{k}{(k^2+k+1)(k^2-k+1)} $$ Expanding into partial fractions, we have: $$ \frac{k}{(k^2+k+1)(k^2-k+1)}= \frac{A}{k^2-k+1} + \frac{B}{k^2+k+1}$$
$$k=A(k^2-k+1)+B(k^2+k+1)$$ $$0k^2+ k^1+0k^0=k(A+B)^2+(-A+B)k+(A+ B)$$ Comparing the coefficients, we have:
$$\begin{cases} A+B=0 \\ -A+B=1 \\ A+B=0\end{cases}$$ $A=\frac{-1}{2}$, $B=\frac{1}{2}$
So $$\frac{k}{(k^2+k+1)(k^2-k+1)}=\frac{1}{2}\frac{-1}{(k^2-k+1)}-\frac{1}{2(k^2+k+1)}$$ $$=- \frac{1}{2} \sum^{50}_{k=1}\left(\frac{k+1}{k^2+k+1}-\frac{k}{k^2-k+1}\right) - \frac{1}{2}\sum^{50}_{k=1}\left(\frac{-k}{k^2+k+1}-\frac{1-k}{k^2-k+1}\right)$$ $$ -\frac{1}{2}\left(\frac{51}{50^2+51}-\frac{1}{1^2}\right)-\frac{1}{2}\left(\frac{-50}{50^2+51}-0\right)= \frac{1275}{2551}$$
I'm right?
You started correctly but made partial fraction wrongly
$$\sum^{50}_{k=1} \frac{k}{(k^2+k+1)(k^2-k+1)} = \frac{1}{2}\sum^{50}_{k=1} \frac{2k}{(k^2+k+1)(k^2-k+1)}$$
Now $2k$ in numerator can be written as $(k^2+k+1)-(k^2-k+1)$
and question will transform into telescoping series
$$\frac{1}{2}\sum^{50}_{k=1} \frac{(k^2+k+1)-(k^2-k+1)}{(k^2+k+1)(k^2-k+1)}$$
$$=\frac{1}{2}\sum^{50}_{k=1}\frac{1}{(k^2-k+1)}-\frac{1}{(k^2+k+1)}$$