Q. If $f(x) = x^{n} + 4$, then the value of $f(1) + \frac{f^{'}(1)}{1!} + \frac{f^{''}(1)}{2!} + ... + \frac{f^{n}(1)}{n!}$ is
A) $2^{n - 1}$
B) $2^{n} + 4$
C) $1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{2!} + ... + \frac{1}{n!}$
D) None of these
Explanation
(b)
$f(1) = 5,f^{'}(x) = nx^{n - 1}so\ f^{'}(1) = n$
$f^{''}(1) = n(n - 1),\ldots f^{n}(1) = 1.2.\ldots.n$
Thus, $f(1) + \frac{f^{'}(1)}{1!} + ... + \frac{f^{n}(1)}{n!}$
$= 5 + \frac{n}{1} + \frac{n(n - 1)}{2!} + ... + \frac{n!}{n!}$
$= (1 + 1)^{n} + 4 = 2^{n} + 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.