Question 90

$$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.......+\frac{1}{n(n+1)}$$= ?

Solution

The given series is: $$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.......+\frac{1}{(n+1)}$$
This can be reframed as: $$\frac{1}{1}n-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}........+\frac{1}{n+1}$$ = $$\frac{1}{1}-\frac{1}{n+1}$$ = $$\frac{n}{n+1}$$
Therefore, the correct option is option D.

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$$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.......+\frac{1}{n(n+1)}$$= ?

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