Question

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

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

Answer 1

The correct option is B $2^n$

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