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JEE Mains · Maths · STD 11 - 8. sequence and series

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 :-

  1. A \(423\)
  2. B \(420\)
  3. C \(421\)
  4. D \(422\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(421\)

Step-by-step Solution

Detailed explanation

\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\right)\left(\mathrm{r}^2-\mathrm{r}+2\right)} \\ & 2…

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