A) \[2n\,(n-1)\]
B) \[2n(n+1)\]
C) \[2{{n}^{2}}(n+1)\]
D) \[2{{n}^{2}}(n-1)\]
E) None of these
Correct Answer: E
Solution :
Given, nth term of a series is \[{{T}_{n}}={{n}^{2}}-2n\] \[Sum=\Sigma {{T}_{n}}=\Sigma ({{n}^{2}}-2n)\] \[=\Sigma {{n}^{2}}-2\Sigma n\] \[=\frac{n(n+1)(2n+1)}{6}-\frac{2n(n+1)}{2}\] \[=n(n+1)\left[ \frac{(2n+1)}{6}-1 \right]\] \[=n(n+1)\left[ \frac{2n+1-6}{6} \right]\] \[=\frac{n(n+1)(2n-5)}{6}\]
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