Sequence And Series Question 130
Question: The sum of the series $ 1.2.3+2.3.4+3.4.5+……. $ to n terms is
[Kurukshetra CEE 1998]
Options:
A) $ n(n+1)(n+2) $
B) $ (n+1)(n+2)(n+3) $
C) $ \frac{1}{4}n(n+1)(n+2)(n+3) $
D) $ \frac{1}{4}(n+1)(n+2)(n+3) $
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Answer:
Correct Answer: C
Solution:
$ T_{n}=n,(n+1),(n+2) $ $ =n,(n^{2}+3n+2)=n^{3}+3n^{2}+2n $
$ \therefore $ $ S_{n}=\Sigma (n^{3})+\Sigma (3n^{2})+\Sigma (2n) $ $ S_{n}={{[ \frac{n,(n+1)}{2} ]}^{2}}+\frac{3.n,(n+1)(2n+1)}{6}+\frac{2.n,(n+1)}{2} $ $ S_{n}=\frac{1}{4}n,(n+1),(n+2)(n+3) $ .
BETA
