Sequence And Series Question 111
Question: The sum of $ (n+1) $ terms of $ \frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+……\text{is } $
[RPET 1999]
Options:
A) $ \frac{n}{n+1} $
B) $ \frac{2n}{n+1} $
C) $ \frac{2}{n,(n+1)} $
D) $ \frac{2,(n+1)}{n+2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ T_{n}=\frac{1}{[ \frac{n(n+1)}{2} ]}=2[ \frac{1}{n}-\frac{1}{n+1} ] $ Put $ n=1,,2,,3,….,(n+1) $ $ T_1=2,[ \frac{1}{1}-\frac{1}{2} ],,T_2=2,[ \frac{1}{2}-\frac{1}{3} ],,…….., $ $ {T_{n+1}}=2[ \frac{1}{n+1}-\frac{1}{n+2} ] $ Hence sum of (n + 1) terms $ =\sum\limits_{k=1}^{n+1}{T_{k}} $
$ \Rightarrow {S_{n+1}}=2[ 1-\frac{1}{n+2} ] $
$ \Rightarrow {S_{n+1,}},=\frac{2(n+1)}{(n+2)} $ .
BETA
