Question

If \[ \sum_{r=1}^{25}\frac{r}{r^4+r^2+1}=\frac{p}{q}, \] where \(p\) and \(q\) are coprime positive integers, then \(p+q\) is equal to:

• A. \(841\)
• B. \(976\)
• C. \(984\)
• D. \(8\)

Hint:

Expressions of the form \(\dfrac{r}{(r^2+r+1)(r^2-r+1)}\) often split into a {difference of two simple fractions}, leading to telescoping sums.

Answer 1

The Correct Option is B

Step 1: Factor the denominator
\[ r^4+r^2+1=(r^2+r+1)(r^2-r+1) \] Hence, \[ \frac{r}{r^4+r^2+1} =\frac{r}{(r^2+r+1)(r^2-r+1)} \]
Step 2: Use partial fractions
Observe that: \[ \frac{r}{(r^2+r+1)(r^2-r+1)} =\frac12\!\left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right) \]
Step 3: Write the series
\[ \sum_{r=1}^{25}\frac{r}{r^4+r^2+1} =\frac12\sum_{r=1}^{25} \left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right) \] This is a telescoping series
.
Step 4: Cancel intermediate terms
All intermediate terms cancel out, leaving: \[ =\frac12\left(\frac{1}{1^2-1+1}-\frac{1}{25^2+25+1}\right) \] \[ =\frac12\left(1-\frac{1}{651}\right) =\frac12\cdot\frac{650}{651} =\frac{325}{651} \]
Step 5: Find \(p+q\)
\[ p=325,\quad q=651 \] \[ p+q=325+651=976 \] \[ \boxed{976} \]