Mathematics - Sequence & Series Question with Solution | TestHub

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