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The value of the sum 12+16+112+120+130+⋯+1n(n+1)\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{n(n+1)}21​+61​+121
Question

The value of the sum 12+16+112+120+130++1n(n+1)\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{n(n+1)}​ is equal to:

A.

1n\frac{1}{n}​​

B.

n(n+1)\frac{n}{(n+1)}​​

C.

1(n+1)\frac{1}{(n+1)}​​

D.

n(n2+1)\frac{n}{(n^2+1)}​​

Correct option is B

Given
Series: 12+16+112+...+1n(n+1)\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + ... + \frac{1}{n(n+1)}​​

Formula Used
1r(r+1)=1r1r+1\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}​​

Solution
The sum can be written as:
(1112)+(1213)+(1314)+...+(1n1n+1)(\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:
112+1213+...1n+1 Sum=11n+1 Sum=n+11n+1=nn+11 - \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)

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