If the sum of the first 20 terms of the series
$\frac{4.1}{4+3.1^2+1^4}+\frac{4.2}{4+3.2^2+2^4}+\frac{4.3}{4+3.3^2+3^4}+\frac{4.4}{4+3.4^2+4^4}+\ldots$
is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m+n$ is equal to :-

Question

Aicharya · Accepted Answer

$\begin{aligned} & \sum_{\mathrm{r}=1}^{20} \frac{4 \mathrm{r}}{4+3 \mathrm{r}^2+\mathrm{r}^4} \ & \sum_{\mathrm{r}=1}^{20} \frac{4 \mathrm{r}}{\left(\mathrm{r}^2+\mathrm{r}+2ight)\left(\mathrm{r}^2-\mathrm{r}+2ight)} \ & 2 \sum_{\mathrm{r}=1}^{20}\left(\frac{1}{\mathrm{r}^2-\mathrm{r}+2}-\frac{1}{\mathrm{r}^2+\mathrm{r}+2}ight) \ & 2\left(\frac{1}{2}-\frac{1}{4}ight) \ & \frac{1}{4}-\frac{1}{8} \ & \frac{1}{8}-\frac{1}{14} \ & \left(\frac{1}{382}-\frac{1}{422}ight) \ & =2\left(\frac{1}{2}-\frac{1}{422}ight) \ & =\frac{420}{422} \ & =\frac{210}{211} \ & 	ext { option }(3)\end{aligned}$ ^

If the sum of the first 20 terms of the series $\frac{4.1}{4+3.1^2+1^4}+\frac{4.2}{4+3.2^2+2^4}+\frac{4…

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