Question

The sum of the series
`1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)`+… to `infty` is
A. 2e
B. 3e
C. 4e
D. 5e

Answer 1

Answer:
We have
`1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)+…underset(n=1)overset(infty)Sigma(n^(3))/(n!)`
Let `n^(3)=a_(0)+a_(1)n+a_(2)n(n-1)+a+_(3)n(n-1)(n-2)`
By comparing the coefficients of like powers of n on both sides we get
Substituting the value in (i) we get
`n^(3)=n+3n(n-1)+n((n-1)(n-2)`
`therefore 1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)+...`
`-underset(n=1)overset(infty)Sigma(n^(3))/(n!)`
`=underset(n=1)overset(infty)Sigma(n+3n(n-1)+n(n-1)(n-2))/(n!)`
=e+3e+e=5e