Sequence And Series Question 152
Question: If $ T_{n}=\frac{3^{n}}{2,(n,!)}-\frac{1}{2,(n,!)}, $ then $ {S_{\infty }}= $
Options:
A) $ \frac{e^{3}-1}{2} $
B) $ \frac{e^{3}-e}{2} $
C) $ \frac{e-3}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Given that $ T_{n}=\frac{1}{2}[ \frac{3^{n}}{n!}-\frac{1}{n!} ] $ Therefore sum of the series $ \sum\limits_{n=1}^{\infty }{T_{n}=\frac{1}{2}[ \sum\limits_{n=1}^{\infty }{,\frac{3^{n}}{n!}-\sum\limits_{n=1}^{\infty }{,\frac{1}{n!}}} ]} $ $ =\frac{1}{2}[ { 1+\frac{3}{1!}+\frac{3^{2}}{2!}+…. }-{ 1+\frac{1}{1!}+\frac{1}{2!}+…. } ] $ $ =\frac{1}{2}(e^{3}-e) $ .
BETA
