A) \[{{n}^{3}}\]
B) \[\frac{1}{3}n\,(n+1)(n+2)\]
C) \[\frac{1}{6}n\,(n+1)(n+2)\]
D) \[\frac{1}{3}n\,(n+1)(2n+1)\]
Correct Answer: B
Solution :
The first factors of the terms of the given series \[1,\ 2,\ 3,\ 4,\ ........n\] and second factors of the terms of the given series \[2,\ 3,\ 4,,\ ........(n+1)\] \[{{n}^{th}}\]term of the given series \[=n(n+1)={{n}^{2}}+n\] Hence sum = \[\Sigma {{n}^{2}}+\Sigma n=\frac{1}{6}n(n+1)(2n+1)+\frac{n}{2}(n+1)\] \[=\frac{1}{6}n(n+1)(2n+1+3)=\frac{1}{3}n(n+1)(n+2)\].
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