Question:
$\frac{1}{3^{2}-1}+\frac{1}{5^{2}-1}+\frac{1}{7^{2}-1}+\ldots+\frac{1}{(201)^{2}-1}$ is equal
to
Correct Option: , 2
Solution:
$T_{n}=\frac{1}{(2 n+1)^{2}-1} \frac{1}{(2 n+2) 2 n}=\frac{1}{4(n)(n+1)}$
$=\frac{(n+1)-n}{4 n(n+1)}=\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$S=\frac{1}{4}\left(1-\frac{1}{101}\right)=\frac{1}{4}\left(\frac{100}{101}\right)=\frac{25}{101}$
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