Answer:
We observe that the successive difference of the terms of the series 3+5+9+15+23+…form an A.P So let its `n^(th)` term be
`t_(n)=an^(2)+bn+c`
Putting n=1,2,3, we get
`t_(1)=a+b+c rarr a+b+c=3`
`t_(2)=4a+2b+crarr4a+2b+c=5`
`t_(3)=9u+3b+crarr9a+3b+c=9`
solving these equation we get
a=1,b=-1 and c =3
`t_(n)=n^(2)-n+3`
`(3)/(1!)+(5)/(2!)+(9)/(3!)+(15)/(4!)`+.....
`=underset(n=1)overset(infty)Sigma (n^(2))/(n!)-underset(n=1)overset(infty)Sigma(n)/(n!)+3underset(n=1)overset(infty)Sigma(1)/(n!)=2e-e+3(e-1)=4e-3`
