Sequence And Series Question 122
Question: $ \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+……..+…….\frac{1}{n.(n+1)} $ equals
[AMU 1995; RPET 1996; UPSEAT 1999, 2001]
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:
$ ( \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} ) $ $ =1-\frac{1}{n+1}=\frac{n}{n+1} $ .
BETA
