Question

If $f\left(x\right)=(1+x{)}^n$, then the value of $f\left(0\right)+f^{\prime }\left(0\right)+\frac{f^{\prime \prime }\left(0\right)}{2!}+\cdots +\frac{f^n\left(0\right)}{n!}$ is

• A. $n$
• B. $2^n$
• C. $2^{n-1}$
• D. None of these.

Answer 1

The correct option is B $2^n$

$f\left(0\right)=1$, $f^{\prime }\left(x\right)=n(1+x{)}^{n-1}$, $f^{\prime \prime }\left(x\right)=n(n-1)(1+x{)}^{n-2}$, $\cdots$ , $f^n\left(x\right)=n(n-1)\cdots 1(1+x{)}^{n-n}$.
So, $f^{\prime }\left(0\right)=n$, $f^{\prime \prime }\left(0\right)=n(n-1)$, $\cdots$, $f^n\left(0\right)=n!$.
Hence the given expression is equal to
$1+n+\frac{n(n-1)}{2!}+\cdots +\frac{n!}{n!}={}^n{C}_0+{}^n{C}_1+{}^n{C}_2+\cdots +{}^n{C}_n=2^n$.