The sum of the series \[{{1}^{2}}.2+{{2}^{2}}.3+{{3}^{2}}.4+........\] to n terms is

\[\frac{n(n+1)}{6}[n(n+1)+(2n+1)]\]

\[\frac{{{n}^{3}}{{(n+1)}^{3}}(2n+1)}{24}\]

\[\frac{n(n+1)(3{{n}^{2}}+7n+2)}{12}\]

\[\frac{n(n+1)}{12}[6n(n+1)+2(2n+1)]\]

The sum of the series \[{{1}^{2}}.2+{{2}^{2}}.3+{{3}^{2}}.4+........\]

1.	The sum of the series \[{{1}^{2}}.2+{{2}^{2}}.3+{{3}^{2}}.4+........\] to n terms is
A.	\[\frac{{{n}^{3}}{{(n+1)}^{3}}(2n+1)}{24}\]
B.	\[\frac{n(n+1)(3{{n}^{2}}+7n+2)}{12}\]
C.	\[\frac{n(n+1)}{6}[n(n+1)+(2n+1)]\]
D.	\[\frac{n(n+1)}{12}[6n(n+1)+2(2n+1)]\]
Answer» C. \[\frac{n(n+1)}{6}[n(n+1)+(2n+1)]\]

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The sum of the series \[{{1}^{2}}.2+{{2}^{2}}.3+{{3}^{2}}.4+........\] to n terms is

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