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 $\frac{\mathrm{m}}{\mathrm{n}}$...

Accepted Answer

$\begin{gathered} {S}_{n}=\sum _{r=1}^{n}\frac{4r}{4+3{r}^{2}+{r}^{4}} \ =2\sum _{r=1}^{n}\frac{2r}{{({r}^{2}+2)}^{2}-{r}^{2}}=2\sum _{r=1}^{n}\frac{({r}^{2}+2+r)-({r}^{2}+2-r)}{({r}^{2}+2+r)({r}^{2}+2-r)} \ =2\sum _{r=1}^{n}(\frac{1}{{r}^{2}+2-r}-\frac{1}{{r}^{2}+2+r}) \ {S}_{20}=2[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{8})+\ldots ] \ \, =2(\frac{1}{2}-\frac{1}{{20}^{2}+2+20}) \ \, =2(\frac{1}{2}-\frac{1}{422}) \ \, =2(\frac{422-2}{422	imes 2})=\frac{420}{422}=\frac{210}{211}=\frac{m}{n} \ m+n=421 \end{gathered}$

Solutions · 1