A) \[\frac{n\,(3n-1)}{2(n)\,!}\]
B) \[\frac{n\,(3n+1)}{2\,(n)\,!}\]
C) \[\frac{n}{2}\frac{3n}{(n)\,!}\]
D) None of these
Correct Answer: B
Solution :
The series is \[\frac{2}{1!}+\frac{(2+5)}{2!}+\frac{\left( 2+5+8 \right)}{3!}+\frac{(2+5+8+11)}{4!}+....\] Hence \[{{T}_{n}}=\frac{(2+5+8+......n\,\text{terms})}{n!}=\frac{\frac{n}{2}[2.2+(n-1)3]}{n!}\] \[{{T}_{n}}=\frac{n\,(3n+1)}{2(n)!}\].
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