A) \[\frac{1}{n(n+1)}\]
B) \[\frac{n}{n+1}\]
C) \[\frac{2n}{n+1}\]
D) \[\frac{2}{n(n+1)}\]
Correct Answer: B
Solution :
| [b] Let \[S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}\] |
| Now, \[{{n}^{th}}\] term of above series \[={{a}_{n}}=\frac{1}{n(n+1)}\] |
| \[\Rightarrow \,\,\,\,{{a}_{n}}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\] (by fraction) |
| Now, \[S=\Sigma {{a}_{n}}=\Sigma \frac{1}{n}-\Sigma \frac{1}{n+1}\] |
| \[=\left( 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n} \right)-\left( \frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}+\frac{1}{n+1} \right)\] |
| \[=1+\left( \frac{1}{2}-\frac{1}{2} \right)+\left( \frac{1}{3}-\frac{1}{3} \right)+.......+\left( \frac{1}{n}-\frac{1}{n} \right)-\frac{1}{n+1}\] |
| \[=1-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\] |
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