Question

If $f\left(x\right)=(1+x{)}^n$ then the value of $f\left(0\right)+f^{\prime }\left(0\right)+...+\frac{f^n\left(0\right)}{n!}$ is

• A. $n$
• B. $2^n$
• C. $2^{n-1}$
• D. $2^{n-2}$

Answer 1

The correct option is B

$2^n$

Given, $f\left(x\right)=(1+x{)}^n$

So, $f^{\prime }\left(x\right)=n(1+x{)}^{n–1}$

$f^{\prime \prime }\left(x\right)=n(n–1)(1+x{)}^{n–2}$

$f^{\prime \prime \prime }\left(x\right)=n(n-1)(n-2)(1+x{)}^{n-3}$

.

.

.

$f^n\left(x\right)=n(n–1)......2.1(1+x{)}^0$

$f\left(0\right)=1,f^{\prime }\left(0\right)=n,f^{\prime \prime }\left(0\right)=n(n–1),f^{\prime \prime \prime }\left(x\right)=n(n-1)(n-2)$

Thus, $f\left(0\right)+f^{\prime }\left(0\right)+...+\frac{f^n\left(0\right)}{n!}$

$=1+n+\frac{n(n-1)}{2!}...\frac{n(n-1)...2.1}{n!}$

$={}^n{C}_0+{}^n{C}_1+{}^n{C}_2+...+{}^n{C}_n=2^n$