Sequence And Series Question 136
Question: The $ n^{th} $ term of series $ \frac{1}{1}+\frac{1+2}{2}+\frac{1+2+3}{3}+……. $ will be
[AMU 1982]
Options:
A) $ \frac{n+1}{2} $
B) $ \frac{n-1}{2} $
C) $ \frac{n^{2}+1}{2} $
D) $ \frac{n^{2}-1}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given series $ \frac{1}{1}+\frac{1+2}{2}+\frac{1+2+3}{3}+…… $ So $ n^{th} $ term of series is given by $ T_{n}=\frac{1+2+3+…….+n}{n}=\frac{\frac{1}{2}n(n+1)}{n}=\frac{n+1}{2} $
BETA
