Sequence And Series Question 162
Question: If the sum of $ 1+\frac{1+2}{2}+\frac{1+2+3}{3}+….. $ to n terms is S, then S is equal to
[Kerala (Engg.) 2002]
Options:
A) $ \frac{n(n+3)}{4} $
B) $ \frac{n(n+2)}{4} $
C) $ \frac{n(n+1),(n+2)}{6} $
D) $ n^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ T_{n}=\frac{1+2+3+…..+n}{n}=\frac{n(n+1)}{2n}=\frac{1}{2}(n+1) $ Hence, $ S=\frac{1}{2}(\Sigma n+n) $ $ =\frac{1}{2}{ \frac{n(n+1)}{2}+n }=\frac{n(n+3)}{4} $ .
BETA
