If the sum of the first 20 terms of the series (4 1)/(4+3 1^2+1^4)+(4 2)/(4+3 2^2+2^4)+(4 …

Question

If the sum of the first 20 terms of the series
$\frac{4 \cdot 1}{4+3 \cdot 1^2+1^4}+\frac{4 \cdot 2}{4+3 \cdot 2^2+2^4}+\frac{4 \cdot 3}{4+3 \cdot 3^2+3^4}+\frac{4 \cdot 4}{4+3 \cdot 4^2+4^4}+\ldots \cdot$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where m and n are coprime, then $\mathrm{m}+\mathrm{n}$ is equal to :

Accepted Answer

$$\begin{aligned}
& S_n=\sum_{r=1}^n \frac{4 r}{4+3 r^2+r^4} \
& =2 \sum_{r=1}^n \frac{2 r}{\left(r^2+2\right)^2-r^2}=2 \sum_{r=1}^n \frac{\left(r^2+2+r\right)-\left(r^2+2-r\right)}{\left(r^2+2+r\right)\left(r^2+2-r\right)} \
& =2 \sum_{r=1}^n\left(\frac{1}{r^2+2-r}-\frac{1}{r^2+2+r}\right) \
& S_{20}=2\left[\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\ldots\right] \
& \quad=2\left(\frac{1}{2}-\frac{1}{20^2+2+20}\right) \
& \quad=2\left(\frac{1}{2}-\frac{1}{422}\right) \
& \quad=2\left(\frac{422-2}{422 \times 2}\right)=\frac{420}{422}=\frac{210}{211}=\frac{m}{n} \
& m+n=421
\end{aligned}$$

Solutions · 1