If r = 1 25 ( r r 4 + r 2 + 1 ) = p q , where p and q are positive integers such that gcd (p,q)=1, then p+q is equal to ________. [2026]

Question

If $\sum_{r = 1}^{25} (\frac{r}{r^{4} + r^{2} + 1}) = \frac{p}{q},$ where p and q are positive integers such that gcd (p,q)=1, then p+q is equal to ________. [2026]

Smart Achievers · Accepted Answer

(976)

$S = \sum \frac{r}{(r^{2} + r + 1) (r^{2} - r + 1)}$

$= \frac{1}{2} \sum_{r = 1}^{25} (\frac{1}{r^{2} - r + 1} - \frac{1}{r^{2} + r + 1})$

$= \frac{1}{2} [(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{7}) + \dots + (\frac{1}{601} - \frac{1}{651})]$

$= \frac{1}{2} [1 - \frac{1}{651}]$

$= \frac{1}{2} [\frac{650}{651}] = \frac{325}{651}$

$\frac{p}{q} = \frac{325}{651} \Rightarrow p + q = 976$

Solution

Q.
If $\sum_{r = 1}^{25} (\frac{r}{r^{4} + r^{2} + 1}) = \frac{p}{q},$ where p and q are positive integers such that gcd (p,q)=1, then p+q is equal to ________. [2026]

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Solution

Q. If ∑r=125(rr4+r2+1)=pq, where p and q are positive integers such that gcd (p,q)=1, then p+q is equal to ________. [2026]

Ans. (976) S=∑r(r2+r+1)(r2-r+1) =12∑r=125(1r2-r+1-1r2+r+1) =12[(11-13)+(13-17)+⋯+(1601-1651)] =12[1-1651] =12[650651]=325651 pq=325651⇒p+q=976

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Q.
If $\sum_{r = 1}^{25} (\frac{r}{r^{4} + r^{2} + 1}) = \frac{p}{q},$ where p and q are positive integers such that gcd (p,q)=1, then p+q is equal to ________. [2026]