Sequence And Series Question 119
Question: The sum of the series $ 1+(1+2)+(1+2+3)+………… $ upto $ n $ terms, will be
[MP PET 1986]
Options:
A) $ n^{2}-2n+6 $
B) $ \frac{n(n+1)(2n-1)}{6} $
C) $ n^{2}+2n+6 $
D) $ \frac{n(n+1)(n+2)}{6} $
Show Answer
Answer:
Correct Answer: D
Solution:
Here $ T_{n}=\frac{n(n+1)}{2} $ Therefore $ S_{n}=\frac{1}{2}{ \Sigma n^{2}+\Sigma n } $ $ =\frac{n(n+1)(n+2)}{6} $ .
BETA
