If the sum of the first 10 terms of the series $\frac{4.1}{1+4.1^4}+\frac{4.2}{1+4.2^4}+\frac{4.3}{1+4.3^4}+\ldots$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $m+n$ is equal to______

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{4 . \mathrm{r}}{1+4 . \mathrm{r}^4} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{4 . \mathrm{r}}{\left(2 \mathrm{r}^2+2 \mathrm{r}+1\right)\left(2 \mathrm{r}^2-2 \mathrm{r}+1\right)} \\ & \mathrm{T}_{\mathrm{r}}=\frac{\left(2 \mathrm{r}^2+2 \mathrm{r}+1\right)-\left(2 \mathrm{r}^2-2 \mathrm{r}+1\right)}{\left(2 \mathrm{r}^2+2 \mathrm{r}+1\right)\left(2 \mathrm{r}^2-2 \mathrm{r}+1\right)}\end{aligned}$
$\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{1}{2 \mathrm{r}^2-2 \mathrm{r}+1}-\frac{1}{2 \mathrm{r}^2+2 \mathrm{r}+1} \\ & \mathrm{~T}_1=\frac{1}{1}-\frac{1}{5} \\ & \mathrm{~T}_2=\frac{1}{5}-\frac{1}{13}\end{aligned}$
$\begin{aligned} & \mathrm{T}_{10}=\frac{1}{181}-\frac{1}{221} \\ & \mathrm{~S}_{10}=1-\frac{1}{221}=\frac{220}{221}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{m}+\mathrm{n}=441\end{aligned}$

If the sum of the first 10 terms of the series $\frac{4.1}{1+4.1^4}+\frac{4.2}{1+4.2^4}+\frac{4.3}{1+4…

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