Explanation
\begin{aligned}
& \sum_{r=1}^{n}T_{r}=\frac{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)}{64} \\
& T_{n}=S_{n}-S_{n-1}=\frac{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)}{64} \\
& =\frac{-\left(2n-3\right)\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}{64} \\
& =\frac{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)\left(2n+5-2n+3\right)}{64} \\
& =\frac{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}{8} \\
& \because\sum_{r=1}^{n}\frac{1}{T_{r}}=8\sum_{r=1}^{n}\frac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)} \\
& =\frac{8}{4}\left(\sum_{r=1}^{n}\frac{1}{\left(2n-1\right)\left(2n+1\right)}-\frac{1}{\left(2n+1\right)\left(2n+3\right)}\right)
\end{aligned}
[1⋅31−3⋅51]+[3⋅51−5⋅71]⋯
<br>=2[(2n−11⋅2n+11−(2n+1)(2n+3)1)]
=2(31−(2n+1)(2n+3)1)
At n→∞ required value =2×31=32
