If the sum of the first 20 terms of the series _r (4r)/(4+3r^2+r^4) is mn, where m,n are c…

Question

If the sum of the first 20 terms of the series $\displaystyle\sum_{r} \frac{4r}{4+3r^2+r^4}$ is $\dfrac{m}{n}$, where $m,n$ are coprime, then $m+n$ is equal to:

Answer: B (421)

Accepted Answer

$4+3r^2+r^4 = (r^2+2)^2-r^2 = (r^2+2-r)(r^2+2+r)$, so
$$\frac{4r}{4+3r^2+r^4} = \frac{2\big[(r^2+2+r)-(r^2+2-r)\big]}{(r^2+2-r)(r^2+2+r)} = 2\left(\frac{1}{r^2-r+2}-\frac{1}{r^2+r+2}\right),$$
a telescoping sum.

$$S_{20} = 2\left(\frac{1}{2}-\frac{1}{422}\right) = \frac{420}{422} = \frac{210}{211}.$$

Since $\gcd(210,211)=1$, $m+n = 421$.

Solutions · 1