Sequence And Series Question 543
Question: $ 1.5+\frac{2.6}{1,!}+\frac{3.7}{2,!}+\frac{4.8}{3,!}+….. $ is equal to
Options:
A) $ 13,e $
B) $ 15,e $
C) $ 9,e+1 $
D) $ 5,e $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 1.5+\frac{2.6}{1!}+\frac{3.7}{2!}+\frac{4.8}{3!}+…. $ $ T_{n}=\frac{n(n+4)}{(n-1)!}=\frac{(n-1)(n+4)}{(n-1)!}+\frac{(n+4)}{(n-1)!} $ $ =\frac{n+4}{(n-2)!}+\frac{1}{(n-2)!}+\frac{5}{(n-1)!} $ $ =\frac{1}{(n-3)!}+\frac{7}{(n-2)!}+\frac{5}{(n-1)!} $ $ {S_{\infty }}=\sum\limits_{n=1}^{\infty }{\frac{1}{(n-3)!}+7\sum\limits_{n=1}^{\infty }{\frac{1}{(n-2)!}+5\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)!}}}} $ $ =e+7e+5e=13e $ .
BETA
