Sequence And Series Question 403
Question: $ \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+….+\frac{1}{n(n+1)} $ equals
Options:
A) $ \frac{1}{n(n+1)} $
B) $ \frac{n}{n+1} $
C) $ \frac{2n}{n+1} $
D) $ \frac{2}{n(n+1)} $
Show Answer
Answer:
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} $ $ =( 1+\frac{1}{2}+\frac{1}{3}+….+\frac{1}{n} )-( \frac{1}{2}+\frac{1}{3}+….+\frac{1}{n}+\frac{1}{n+1} ) $ $ =1+( \frac{1}{2}-\frac{1}{2} )+( \frac{1}{3}-\frac{1}{3} )+…….+( \frac{1}{n}-\frac{1}{n} )-\frac{1}{n+1} $ $ =1-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1} $
BETA
