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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Sequences and Series

If ∑_ r=1 ^ 25 ( r r^ 4 +r^ 2 +1 )=(p)/(q) , where p and q are...

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

Correct answer

0

Step-by-step solution

S=∑r(r2+r+1)(r2−r+1)S = \sum \dfrac{r}{(r^2 + r + 1)(r^2 - r + 1)}
=12∑r=125(1r2−r+1−1r2+r+1)= \dfrac{1}{2} \sum_{r=1}^{25} \left( \dfrac{1}{r^2 - r + 1} - \dfrac{1}{r^2 + r + 1} \right)
=12[(11−13)+(13−17)+…+(1601−1651)]= \dfrac{1}{2} \left[ \left( \dfrac{1}{1} - \dfrac{1}{3} \right) + \left( \dfrac{1}{3} - \dfrac{1}{7} \right) + \ldots + \left( \dfrac{1}{601} - \dfrac{1}{651} \right) \right]
=12[11−1651]= \dfrac{1}{2} \left[ \dfrac{1}{1} - \dfrac{1}{651} \right]
=12[650651]=325651= \dfrac{1}{2} \left[ \dfrac{650}{651} \right] = \dfrac{325}{651}
pq=325651⇒p+q=976\dfrac{p}{q} = \dfrac{325}{651} \Rightarrow p + q = 976
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