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Sum of k/(k^4 + k^2 + 1)

The same question, asked 3 times: 26 Jul 2022*, 1 Feb 2023*, 28 Jan 2026 S2. Every wording below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    26 Jul 2022

    174
    If ∑k=110kk4+k2+1=mn\sum_{k = 1}^{10} \frac{k}{k^{4}+k^{2}+ 1}=\frac{m}{n}, where mm and nn are coprime, then m+nm + n is equal to

  2. Question 2

    1 Feb 2023

    71
    The sum to 10 terms of the series 11+12+14+21+22+24+31+32+34+⋯\frac{1}{1 +1^{2}+1^{4}}+\frac{2}{1 +2^{2}+2^{4}}+\frac{3}{1 +3^{2}+3^{4}}+ \cdots. is

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  3. Question 3

    28 Jan 2026 S2

    71
    If ∑r=125(rr4+r2+1)=pq\sum_{r = 1}^{25} \left( \frac{r}{r^{4} + r^{2} + 1} \right) = \frac{p}{q}, where p and q are positive integers such that gcd⁡(p,q)=1\gcd(p,q) = 1, then p+qp + q is equal to ............ .