Question ID: #972
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 ________.
Solution:
We need to evaluate the sum:
$$S = \sum_{r=1}^{25} \frac{r}{r^4+r^2+1}$$
Factorize the denominator $r^4+r^2+1$:
$$r^4+r^2+1 = (r^2+1)^2 – r^2 = (r^2-r+1)(r^2+r+1)$$
Rewrite the general term using partial fractions:
$$\frac{r}{(r^2-r+1)(r^2+r+1)} = \frac{1}{2} \left[ \frac{1}{r^2-r+1} – \frac{1}{r^2+r+1} \right]$$
Now, expand the sum from $r=1$ to $r=25$:
$$S = \frac{1}{2} \left[ \left(\frac{1}{1^2-1+1} – \frac{1}{1^2+1+1}\right) + \left(\frac{1}{2^2-2+1} – \frac{1}{2^2+2+1}\right) + \dots + \left(\frac{1}{25^2-25+1} – \frac{1}{25^2+25+1}\right) \right]$$
Evaluate the terms to observe the telescopic cancellation:
$$S = \frac{1}{2} \left[ \left(\frac{1}{1} – \frac{1}{3}\right) + \left(\frac{1}{3} – \frac{1}{7}\right) + \left(\frac{1}{7} – \frac{1}{13}\right) + \dots + \left(\frac{1}{601} – \frac{1}{651}\right) \right]$$
All intermediate terms cancel out, leaving only the first and the last terms:
$$S = \frac{1}{2} \left[ 1 – \frac{1}{651} \right]$$
$$S = \frac{1}{2} \left[ \frac{650}{651} \right] = \frac{325}{651}$$
We are given $S = \frac{p}{q}$ and $\gcd(p,q)=1$.
Here, $p = 325$ and $q = 651$. They have no common factors.
Calculate $p+q$:
$$p+q = 325 + 651 = 976$$
Ans. 976
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