Discussion

Explanation:

Let S =  $\frac{1}{1\times 3}$ + $\frac{1}{3\times 5}$ + $\frac{1}{5\times 7}$ + ...

∴ S = $\frac{1}{2}\left[\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...\right]$

Now, the first term $\frac{1}{1\times 3}$ = $\frac{3-1}{1\times 3}$ = $\frac{1}{1}$ - $\frac{1}{3}$

Similarly, the n^th term which is $\frac{2}{(2\mathrm{n}-1)\times (2\mathrm{n}+1)}$ can be writtern as $\frac{1}{(2\mathrm{n}-1)}$ - $\frac{1}{(2\mathrm{n}+1)}$

⇒ S = $\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\right]$ 

⇒ S = $\frac{1}{2}\left[1-\frac{1}{101}\right]$ = $\frac{50}{101}$

Hence, option (b).

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