Sequence And Series Question 129
Question: The sum of the series $ 1^{2}.2+2^{2}.3+3^{2}.4+…….. $ to n terms is
Options:
A) $ \frac{n^{3}{{(n+1)}^{3}}(2n+1)}{24} $
B) $ \frac{n(n+1)(3n^{2}+7n+2)}{12} $
C) $ \frac{n(n+1)}{6}[n(n+1)+(2n+1)] $
D) $ \frac{n(n+1)}{12}[6n(n+1)+2(2n+1)] $
Show Answer
Answer:
Correct Answer: B
Solution:
$ T_{n}=n^{2}(n+1)=n^{3}+n^{2} $ $ S_{n}=\Sigma T_{n}=\Sigma n^{3}+\Sigma n^{2} $ $ ={{[ \frac{n,(n+1)}{2} ]}^{2}}+\frac{n,(n+1),(2n+1)}{6} $ $ =\frac{n,(n+1)}{2}[ \frac{n,(n+1)}{2}+\frac{2n+1}{3} ] $ $ =\frac{n,(n+1)(3n^{2}+7n+2)}{12} $ .
BETA
