JEE Main 2025 Jan 22 Shift 1, Mathematics Q10: Summation of Series

Write $S_n=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$. Then $T_n=S_n-S_{n-1}$. Factoring out the common cubic:

$T_n=\frac{(2n-1)(2n+1)(2n+3)}{64}\cdot \left[(2n+5)-(2n-3)\right]=\frac{(2n-1)(2n+1)(2n+3)}{8}$.

So $\frac{1}{T_n}=\frac{8}{(2n-1)(2n+1)(2n+3)}$. Partial fractions give

$\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{4}\left[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}\right]$.

Hence $\frac{1}{T_n}=2\left[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}\right]$, a telescoping series.

$\sum _{r=1}^{\infty }\frac{1}{T_r}=2\cdot \frac{1}{1\cdot 3}=\frac{2}{3}$.