If _k=1^10(k)/(k^4+k^2+1) = (m)/(n), where m,n are coprime, then m+n is equal to ________.

Question

If $\displaystyle\sum_{k=1}^{10}\frac{k}{k^4+k^2+1} = \frac{m}{n}$, where $m,n$ are coprime, then $m+n$ is equal to \_\_\_\_\_\_\_\_.

Answer: 166

Accepted Answer

As before, $\dfrac{k}{k^4+k^2+1} = \dfrac12\left[\dfrac{1}{k^2-k+1}-\dfrac{1}{k^2+k+1}\right]$, telescoping to

$$\sum_{k=1}^{10} = \frac12\left(1-\frac{1}{111}\right) = \frac{55}{111} \implies m+n = 166.$$

Solutions · 1