Discussion

Explanation:

Let S =  $\frac{1}{1\times 2\times 3}$ + $\frac{1}{2\times 3\times 4}$ + $\frac{1}{3\times 4\times 5}$ + ...

Now, the first term $\frac{1}{1\times 2\times 3}$ = $\frac{1}{2}\left(\frac{3-1}{1\times 2\times 3}\right)$ = $\frac{1}{2}\left(\frac{1}{1\times 2}-\frac{1}{2\times 3}\right)$

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

⇒ S = $\frac{1}{2}\left(\frac{1}{1\times 2}-\frac{1}{2\times 3}\right)$ + $\frac{1}{2}\left(\frac{1}{2\times 3}-\frac{1}{3\times 4}\right)$ + ... + $\frac{1}{2}\left(\frac{1}{49\times 50}-\frac{1}{50\times 51}\right)$ + $\frac{1}{2}\left(\frac{1}{50\times 51}-\frac{1}{51\times 52}\right)$

⇒ S = $\frac{1}{2}\left(\frac{1}{1\times 2}-\frac{1}{51\times 52}\right)$

Hence, option (b).

» Your doubt will be displayed only after approval.