If sum_{r=1}^{n} T_{r} = frac{(2n-1)(2n+1)(2n | vedclass

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(C) Given $S_{n} = \sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$.$T_{n} = S_{n} - S_{n-1} = \frac{(2n-1)(2n+1)(2n+3)(2n+5) - (2n-3)(2n-1

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$\frac{2}{3}$

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(C) Given $S_{n} = \sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$.$T_{n} = S_{n} - S_{n-1} = \frac{(2n-1)(2n+1)(2n+3)(2n+5) - (2n-3)(2n-1)(2n+1)(2n+3)}{64}$.$T_{n} = \frac{(2n-1)(2n+1)(2n+3) [ (2n+5) - (2n-3) ]}{64} = \frac{(2n-1)(2n+1)(2n+3) 	imes 8}{64} = \frac{(2n-1)(2n+1)(2n+3)}{8}$.Thus,$\frac{1}{T_{r}} = \frac{8}{(2r-1)(2r+1)(2r+3)}$.Using partial fractions: $\frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{1}{4} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} ight)$.So,$\sum_{r=1}^{n} \frac{1}{T_{r}} = 8 	imes \frac{1}{4} \sum_{r=1}^{n} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} ight) = 2 \left( \frac{1}{1 	imes 3} - \frac{1}{(2n+1)(2n+3)} ight)$.Taking the limit as $n ightarrow \infty$,the second term vanishes.Result $= 2 	imes \frac{1}{3} = \frac{2}{3}$.

If $\sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$,then $\lim_{n \rightarrow \infty} \sum_{r=1}^{n} \left(\frac{1}{T_{r}}\right)$ is equal to :

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