Question

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

Hint:

Whenever you see rational expressions involving consecutive quadratic factors, always try partial fractions to look for telescoping.

Answer 1

Correct Answer: 976

To solve the problem, let's evaluate the series: \[\sum_{r=1}^{25}\left(\frac{r}{r^4+r^2+1}\right)=\frac{p}{q}\] where \( \gcd(p,q)=1 \). First, simplify the expression inside the series. Consider the denominator: \(r^4+r^2+1\). Notice that:
\[r^4+r^2+1 = (r^2+1)^2 - r^2 = (r^2+r+1)(r^2-r+1)\]

This factorization helps rewrite our term as:
\[\frac{r}{r^4+r^2+1}=\frac{r}{(r^2+r+1)(r^2-r+1)}\]
Observe that the expression can be broken into partial fractions:
\[\frac{r}{(r^2+r+1)(r^2-r+1)} = \frac{1}{2}\left(\frac{1}{r^2-r+1} - \frac{1}{r^2+r+1}\right)\]
Therefore, the series becomes telescopic:
\[\sum_{r=1}^{25}\frac{1}{2}\left(\frac{1}{r^2-r+1} - \frac{1}{r^2+r+1}\right)\]
When expanded and simplified from \(r=1\) to \(r=25\), we find that most terms cancel out, leaving:
\[\frac{1}{2}\left(1 - \frac{1}{26^2-26+1}\right)\]
Compute the term:
\[26^2 - 26 + 1 = 676 - 26 + 1 = 651\]
The final expression is:
\[\frac{1}{2}\left(1 - \frac{1}{651}\right) = \frac{1}{2}\left(\frac{651 - 1}{651}\right) = \frac{1}{2}\cdot\frac{650}{651} = \frac{325}{651}\]
Notice \(\gcd(325,651)=1\). Thus, \(\frac{p}{q}=\frac{325}{651}\) and \(p+q = 325 + 651 = 976\).
Finally, 976 fits within the expected range [976,976]. Therefore, the result of \(p+q\) is 976.