Q. If

$\displaystyle \sum_{r=1}^{25}\left(\frac{r}{r^4+r^2+1}\right)=\frac{p}{q},$

where $p$ and $q$ are positive integers such that $\gcd(p,q)=1$, then $p+q$ is equal to ______.

Correct Answer: 976

Explanation

We simplify the denominator:

$\displaystyle r^4+r^2+1=(r^2-r+1)(r^2+r+1)$

Now perform partial fraction decomposition:

$\displaystyle \frac{r}{r^4+r^2+1} =\frac{1}{2}\left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right)$

So the summation becomes:

$\displaystyle \sum_{r=1}^{25}\frac{r}{r^4+r^2+1} =\frac{1}{2}\sum_{r=1}^{25}\left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right)$

This is a telescoping series.

Writing first and last few terms:

$\displaystyle \frac{1}{2}\left[ \left(\frac{1}{1}-\frac{1}{3}\right) +\left(\frac{1}{3}-\frac{1}{7}\right) +\cdots +\left(\frac{1}{601}-\frac{1}{651}\right) \right]$

All intermediate terms cancel.

$\displaystyle =\frac{1}{2}\left(1-\frac{1}{651}\right) =\frac{1}{2}\cdot\frac{650}{651} =\frac{325}{651}$

Here,

$p=325,\quad q=651$

Therefore,

$p+q=325+651=976$

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