मान लीजिए x_{n}=left(1-frac{1}{3}right)^{2}le | vedclass

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(B) सामान्य पद $1 - \frac{1}{\frac{n(n+1)}{2}} = 1 - \frac{2}{n(n+1)} = \frac{n^2+n-2}{n(n+1)} = \frac{(n+2)(n-1)}{n(n+1)}$ है।$x_n = \left[ \prod_{k=

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$1/9$

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(B) सामान्य पद $1 - \frac{1}{\frac{n(n+1)}{2}} = 1 - \frac{2}{n(n+1)} = \frac{n^2+n-2}{n(n+1)} = \frac{(n+2)(n-1)}{n(n+1)}$ है।$x_n = \left[ \prod_{k=2}^{n} \frac{(k+2)(k-1)}{k(k+1)} \right]^2$.$x_n = \left[ \left( \prod_{k=2}^{n} \frac{k+2}{k+1} \right) \left( \prod_{k=2}^{n} \frac{k-1}{k} \right) \right]^2$.गुणनफल की गणना करने पर:$\prod_{k=2}^{n} \frac{k+2}{k+1} = \frac{4}{3} \cdot \frac{5}{4} \cdot \dots \cdot \frac{n+2}{n+1} = \frac{n+2}{3}$.$\prod_{k=2}^{n} \frac{k-1}{k} = \frac{1}{2} \cdot \frac{2}{3} \cdot \dots \cdot \frac{n-1}{n} = \frac{1}{n}$.अतः, $x_n = \left( \frac{n+2}{3} \cdot \frac{1}{n} \right)^2 = \frac{1}{9} \left( \frac{n+2}{n} \right)^2 = \frac{1}{9} \left( 1 + \frac{2}{n} \right)^2$.$n \rightarrow \infty$ पर सीमा लेने पर, $\lim_{n \rightarrow \infty} x_n = \frac{1}{9} (1+0)^2 = \frac{1}{9}$.

मान लीजिए $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} \ldots \left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2}, n \geq 2$ है। तो, $\lim _{n \rightarrow \infty} x_{n}$ का मान ज्ञात कीजिए।

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