Sum of series `9/(1!)+19/(2!)+35/(3!)+57/(4!)+...` (A) `7e-3` (B) `12e-5` (C) `16e-5` (D) none

Options

A. `16e-5`
B. `7e-3`
C. `12e-5`
D. `11e-5`

Explanation

We observe that the successive difference of the terms of the sequence 9,19,35,57,85…are in A.P.So let its `n^(th)` term be
`t_(n)=an^(2)+bn+c`
Putting n=1,2,3 we get
a+b+c=9,4a+2b+c=19+3b+c=35
solving these equation we get
a=3,b=1 and c=5
`therefore t_(n)=3xn^(2)+n+5`
Hence
`(9)/(1!)+(19)/(2!)+(35)/(3!)+...=underset(n=1)overset(infty)Sigma(3n^(2)+n+5)/(n!)`
`rarr (9)/(1!)+(19)/(2!)+(35)/(3!)+....underset(n=1)overset(infty)Sigma (n^(2))/(n!)+underset(n=1)overset(infty)Sigma(1)/(n!)`
`rarr(9)/(1!)+(19)/(2!)+(35)/(3!)+...=3xx2e+e+5(e-1)=12e-5`