If `f(x)=(4+x)^(n),"n" epsilonN` and `f^(r)(0)` represents the `r^(th)` derivative of `f(x)` at `x=0`, then the value of `sum_(r=0)^(oo)((f^(r)(0)))/(r!)` is equal to

Options

A. `2^(n)`
B. `e^(n)`
C. `5^(n)`
D. `4^(n)`

Answer

C

Explanation

`f^(')(x)=n(4+_x)^(n-1),f^(")=n(n-1).(4+x)^(n-2)`
If we keep on differentiating, we get
`f^(r)(0)=(n!)/((n-r)!).4^(n-r),rlen`
and `f^(r)(0)=0` for `rgtn`
`sum_(t=0)^(oo)(f^(r)(0))/(r!)=sum_(r=0)^(n) .^(n)C_(r).4^(n-r)=(1+4)^(n)=5^(n)`