Question

If
$\sum_{k=1}^{10} \frac{k}{k^4 + k^2 + 1} = \frac{m}{n}$
where m and n are co-prime, then m + n is equal to

Answer 1

The correct answer is 166
$\sum_{k=1}^{10} \frac{k}{k^4 + k^2 + 1}$
$=\frac{1}{2} \left[ \sum_{k=1}^{10} \left( \frac{1}{k^2 - k + 1} - \frac{1}{k^2 + k + 1} \right) \right]$
$=\frac{1}{2} \left[ 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{13} + \ldots + \frac{1}{91} - \frac{1}{111} \right]$
$\frac{1}{2} \left[ 1 - \frac{1}{111} \right] = \frac{110}{2 \cdot 111} = \frac{55}{111} = \frac{m}{n}$
∴ m + n = 55 + 111 = 166