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

Step 1: Simplify the general term \[ r^4+r^2+1=(r^2+r+1)(r^2-r+1) \] Now decompose: \[ \frac{r}{r^4+r^2+1} =\frac{1}{2}\left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right) \]
Step 2: Use telescoping nature \[ \sum_{r=1}^{25}\frac{r}{r^4+r^2+1} =\frac12\sum_{r=1}^{25} \left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right) \] Write initial and final terms explicitly: \[ =\frac12\left( \frac{1}{1}-\frac{1}{3} +\frac{1}{3}-\frac{1}{7} +\cdots +\frac{1}{601}-\frac{1}{651} \right) \] All intermediate terms cancel. \[ =\frac12\left(1-\frac{1}{651}\right) =\frac12\cdot\frac{650}{651} =\frac{325}{651} \]
Step 3: Compute \(p+q\) \[ p=325,\ q=651,\ \gcd(325,651)=1 \] \[ p+q=976 \] Final Answer: \[ \boxed{976} \]