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

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

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$\frac{2}{3}$

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(C) दिया गया है $S_{n} = \sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$.$T_{n} = S_{n} - S_{n-1} = \frac{(2n-1)(2n+1)(2n+3)(2n+5) - (2n-3)(2n-1)(2n+1)(2n+3)}{64}$.$T_{n} = \frac{(2n-1)(2n+1)(2n+3) [ (2n+5) - (2n-3) ]}{64} = \frac{(2n-1)(2n+1)(2n+3) 	imes 8}{64} = \frac{(2n-1)(2n+1)(2n+3)}{8}$.अतः,$\frac{1}{T_{r}} = \frac{8}{(2r-1)(2r+1)(2r+3)}$.आंशिक भिन्नों का उपयोग करने पर: $\frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{1}{4} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} ight)$.इसलिए,$\sum_{r=1}^{n} \frac{1}{T_{r}} = 8 	imes \frac{1}{4} \sum_{r=1}^{n} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} ight) = 2 \left( \frac{1}{1 	imes 3} - \frac{1}{(2n+1)(2n+3)} ight)$.$n ightarrow \infty$ पर सीमा लेने पर,दूसरा पद शून्य हो जाएगा।परिणाम $= 2 	imes \frac{1}{3} = \frac{2}{3}$.

यदि $\sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$ है,तो $\lim_{n \rightarrow \infty} \sum_{r=1}^{n} \left(\frac{1}{T_{r}}\right)$ का मान ज्ञात कीजिए:

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