Sequence And Series Question 153
Question: If the $ n^{th} $ term of a series be $ 3+n,(n-1) $ , then the sum of $ n $ terms of the series is
Options:
A) $ \frac{n^{2}+n}{3} $
B) $ \frac{n^{3}+8n}{3} $
C) $ \frac{n^{2}+8n}{5} $
D) $ \frac{n^{2}-8n}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
Here, $ T_{n}=3+n(n-1)=3+n^{2}-n $ Now sum $ S=\Sigma T_{n}=\Sigma (3+n^{2}-n) $ $ =3n+\frac{1}{6}n(n+1)(2n+1)-\frac{n(n+1)}{2} $ $ =\frac{1}{6}n(n+1)[2n+1-3]+3n=\frac{n^{3}+8n}{3} $ .
BETA
