If ∑ Tr=(2n-1)(2n+1)(2n+3)(2n+5)/64 n ∈ to (r=1, n), then

Question

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

(1) 1

(2) 0

(3) \(\frac{2}{3}\)

(4) \(\frac{1}{3}\)

Answer 1

Correct option is (3) \(\frac{2}{3}\) 

\(\mathrm{T}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}\)

\(\Rightarrow \mathrm{T}_{\mathrm{n}}=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)\)

\(\Rightarrow \frac{1}{\mathrm{~T}_{\mathrm{n}}}=\frac{8}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)}\)

\(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{T_{r}}=\lim _{n \rightarrow \infty} 8 \sum_{r=1}^{n} \frac{1}{(2 n-1)(2 n+1)(2 n+3)}\)

\(=\lim _{n \rightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 n-1)(2 n+1)}-\frac{1}{(2 n+1)(2 n+3)}\right)\)

\(=\lim _{n \rightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\ldots\right]\)

\(=\frac{2}{3}\)