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Sum of n/(n^4 + a n^2 + b) by splitting into a difference

This type of question, asked 8 times: 27 Jun 2022, 26 Jul 2022, 1 Feb 2023, 31 Jan 2024, 2 Apr 2025, 4 Apr 2025, 28 Jan 2026 S2, 5 Apr 2026 S2. All 8 questions below, oldest first.

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  1. Question 1

    27 Jun 2022

    87
    If the sum of the first ten terms of the series 15+265+3325+41025+52501+…\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+ \ldots is mn\frac{m}{n}, where mm and nn are co-prime numbers, then m+nm + n is equal to

  2. Question 2

    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

  3. Question 3

    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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  4. Question 4

    31 Jan 2024

    73
    The sum of the series 11−3⋅12+14+\frac{1}{1 - 3 \cdot1^{2}+1^{4}}+ 21−3⋅22+24+31−3⋅32+34+….\frac{2}{1 - 3 \cdot2^{2}+2^{4}}+\frac{3}{1 - 3 \cdot3^{2}+3^{4}}+ \ldots.. up to 10 terms is

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  5. Question 5

    2 Apr 2025

    147
    If the sum of the first 10 terms of the series 4.11+4.14+4.21+4.24+4.31+4.34+…\frac{4.1}{1 + {4.1}^{4}} + \frac{4.2}{1 + {4.2}^{4}} + \frac{4.3}{1 + {4.3}^{4}} + \ldots is mn\frac{m}{n}, where gcd(m,n)=1gcd(m,n) = 1, then m+nm + n is equal to.

  6. Question 6

    4 Apr 2025

    137
    If the sum of the first 20 terms of the series
    4.14+3.12+14+4.24+3.22+24+\frac{4.1}{4 +3.1^{2}+1^{4}}+\frac{4.2}{4 +3.2^{2}+2^{4}}+
    4.34+3.32+34+4.44+3.42+44+….\frac{4.3}{4 +3.3^{2}+3^{4}}+\frac{4.4}{4 +3.4^{2}+4^{4}}+ \ldots.
    is mn\frac{m}{n}, where m and n are coprime, then m+nm + n is equal to :-

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  7. Question 7

    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 ............ .

  8. Question 8

    5 Apr 2026 S2

    55
    If the sum of the first 10 terms of the series 11+14×4+21+24×4+31+34×4+41+44×4+…\frac{1}{1 +1^{4}\times 4}+\frac{2}{1 +2^{4}\times 4}+\frac{3}{1 +3^{4}\times 4}+\frac{4}{1 +4^{4}\times 4}+ \ldots is mn,gcd(m,n)=1\frac{m}{n},gcd(m,n) = 1, then m+nm + n is equal to :

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