frac{1}{3^{2}-1}+frac{1}{5^{2}-1}+frac{1}{7^{ | vedclass

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(B) The general term of the series is $T_n = \frac{1}{(2n+1)^2 - 1}$ for $n = 1, 2, \ldots, 100$.Simplifying the denominator: $(2n+1)^2 - 1 = (2n+1-1)

Vedclass · Accepted Answer

$\frac{25}{101}$

Vedclass · Answer

(B) The general term of the series is $T_n = \frac{1}{(2n+1)^2 - 1}$ for $n = 1, 2, \ldots, 100$.Simplifying the denominator: $(2n+1)^2 - 1 = (2n+1-1)(2n+1+1) = (2n)(2n+2) = 4n(n+1)$.Thus,$T_n = \frac{1}{4n(n+1)} = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.The sum $S = \sum_{n=1}^{100} T_n = \frac{1}{4} \sum_{n=1}^{100} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.This is a telescoping series: $S = \frac{1}{4} \left( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \ldots + (\frac{1}{100} - \frac{1}{101}) \right)$.$S = \frac{1}{4} \left( 1 - \frac{1}{101} \right) = \frac{1}{4} \left( \frac{100}{101} \right) = \frac{25}{101}$.

$\frac{1}{3^{2}-1}+\frac{1}{5^{2}-1}+\frac{1}{7^{2}-1}+\ldots+\frac{1}{(201)^{2}-1}$ is equal to

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