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:
$\mathrm{T}_{\mathrm{n}}=\frac{1}{(2 \mathrm{n}+1)^{2}-1} \frac{1}{(2 \mathrm{n}+2) 2 \mathrm{n}}=\frac{1}{4(\mathrm{n})(\mathrm{n}+1)}$
$=\frac{(\mathrm{n}+1)-\mathrm{n}}{4 \mathrm{n}(\mathrm{n}+1)}=\frac{1}{4}\left(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\right)$
$\mathrm{S}=\frac{1}{4}\left(1-\frac{1}{101}\right)=\frac{1}{4}\left(\frac{100}{101}\right)=\frac{25}{101}$
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