A) \[\frac{n(3n-1)}{2}\]
B) \[\frac{n(3n+1)}{2}\]
C) \[n(3n+2)\]
D) \[\frac{n(3n+1)}{4}\]
Correct Answer: A
Solution :
\[\sum\limits_{i=1}^{n}{{}}=3\sum\limits_{i=1}^{n}{i}-2\sum\limits_{i=1}^{n}{1}=3\frac{n\,(n+1)}{2}-2n=\frac{n\,(3n-1)}{2}\].
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