If \[\sum\limits_{i=1}^{n}{i=\frac{n(n+1)}{2}}\], then \[\sum\limits_{i=1}^{n}{(3i-2)=}\]

If \[\sum\limits_{i=1}^{n}{i=\frac{n(n+1)}{2}}\], then \[\sum\limits

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

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If \[\sum\limits_{i=1}^{n}{i=\frac{n(n+1)}{2}}\], then \[\sum\limits_{i=1}^{n}{(3i-2)=}\]

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