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Q)

Find $\frac{1}{3}+\frac{1}{15}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}$ +....upto n terms

$\begin{array}{1 1} \large\frac{4}{2n-1} \\ \frac{1}{2n+1} \\ \frac{1}{n+1} \\ \frac{n}{2n+1} \end{array}$

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A)
Answer : (d) $\;\frac{2}{2n+1}$
Explanation : $\;a_{r}=\frac{1}{(2r)^2-1}$
$=\large\frac{1}{(2r-1)(2r+1)}$
$=\frac{1}{2}\;[\large\frac{1}{2r-1}-\large\frac{1}{2r+1}]$
$\sum_{r=1}^n\;a_{r}=\frac{1}{2}\;\sum_{r=1}^{n}\;[\large\frac{1}{2r-1}-\large\frac{1}{2r+1}]$
$=\frac{1}{2}\;[1-\large\frac{1}{2n+1}]$
$=\large\frac{n}{2n+1}\;.$
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