Question

If $f(x)=x^n+4$, then the value of $f(1)+\frac{f^{\prime}(1)}{1 !}+\frac{f^{\prime \prime}(1)}{2 !}+\frac{f^{\prime \prime \prime}(1)}{3 !}+\ldots+\frac{f^n(1)}{n !}$ is

• A. $2^{n-1}$
• B. $2^n+4$
• C. $1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots+\frac{1}{n !}$
• D. None of these

Answer 1

Answer: (B)

$f(1)=5, f^{\prime}(x)=n x^{n-1}, f^{\prime \prime}(x)=n(n-1) x^{n-2}$
So, $ f^{\prime}(1)=n$
$f^{\prime \prime}(1)=n(n-1) \ldots f^n(1)=12 \ldots n$
Thus, $f(1)+\frac{f^{\prime}(1)}{1 !}+\ldots+\frac{f^n(1)}{n !}$
$ =5+\frac{n}{1}+\frac{n(n-1)}{2 !}+\ldots+\frac{n !}{n !} $
$ =(1+1)^n+4=2^n+4$