The value of the sum $$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{n(n+1)}$$​ is equal to:

Correct option is B

Given
Series: $$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + ... + \frac{1}{n(n+1)}$$​​

Formula Used
​$$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$$​​

Solution
The sum can be written as:
​$$(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{n} - \frac{1}{n+1})$$​​
All intermediate terms cancel out:
​$$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... - \frac{1}{n+1} \\\ \\Sum = 1 - \frac{1}{n+1} \\\ \\Sum = \frac{n+1-1}{n+1} = \frac{n}{n+1}$$​​

Final Answer
So the correct answer is (b)