Discussion

Explanation:

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

Now, the first term $\frac{1}{1\times 2}$ = $\frac{2-1}{1\times 2}$ = $\frac{1}{1}$ - $\frac{1}{2}$

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

⇒ S = $\left(\frac{1}{1}-\frac{1}{2}\right)$ + $\left(\frac{1}{2}-\frac{1}{3}\right)$ + $\left(\frac{1}{3}-\frac{1}{4}\right)$ .. + $\left(\frac{1}{48}-\frac{1}{49}\right)$ + $\left(\frac{1}{49}-\frac{1}{50}\right)$ + $\left(\frac{1}{50}-\frac{1}{51}\right)$

⇒ S = 1 - $\frac{1}{51}$ = $\frac{50}{51}$

Hence, option (c).

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