Prove the following
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

  1. (1) $\frac{101}{404}$

  2. (2) $\frac{25}{101}$

  3. (3) $\frac{101}{408}$

  4. (4) $\frac{99}{400}$


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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