If _r=1^25 (r)/(r^4+r^2+1) = (p)/(q), where p, q are positive integers such that (p,q)=1, …

Question

If $\displaystyle\sum_{r=1}^{25} \frac{r}{r^4+r^2+1} = \frac{p}{q}$, where $p, q$ are positive integers such that $\gcd(p,q)=1$, then $p+q$ is equal to \_\_\_\_\_\_\_\_.

Answer: 976

Accepted Answer

$$r^4+r^2+1 = (r^2+1)^2 - r^2 = (r^2-r+1)(r^2+r+1).$$

So the general term is
$$T_r = \frac{r}{(r^2-r+1)(r^2+r+1)} = \frac12\left[\frac{1}{r^2-r+1} - \frac{1}{r^2+r+1}\right],$$
since $(r^2+r+1)-(r^2-r+1) = 2r$.

Let $f(r) = \dfrac{1}{r^2-r+1}$. Then $f(r+1) = \dfrac{1}{r^2+r+1}$, so $T_r = \tfrac12[f(r)-f(r+1)]$ --- a telescoping sum:

$$\sum_{r=1}^{25} T_r = \frac12\big[f(1)-f(26)\big] = \frac12\left[1 - \frac{1}{625+25+1}\right] = \frac12\left(\frac{650}{651}\right) = \frac{325}{651}.$$

Since $325 = 5^2\cdot13$ and $651 = 3\cdot7\cdot31$, $\gcd(325,651)=1$, so $p=325$, $q=651$.

$p+q = 325+651 = 976.$

Solutions · 1