Question

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

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

Answer 1

The correct option is B $2^n$

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

$f\left(0\right)=1$
$f^{\prime }\left(x\right)=n(1+x{)}^{n-1}$
$f^{\prime }\left(0\right)=n$
$f^{\prime \prime }\left(x\right)=n(n-1)(1+x{)}^{n-2}$
$f^{\prime \prime }\left(0\right)=n(n-1)$
$f^{\prime \prime \prime }\left(x\right)=n(n-1)(n-2)(1+x{)}^{n-3}$
$f^{\prime \prime \prime }\left(0\right)=n(n-1)(n-2)$

$f\left(0\right)+f^{\prime }\left(0\right)+\frac{f^{\prime \prime }\left(0\right)}{2!}+\frac{f^{\prime \prime \prime }\left(0\right)}{3!}........\frac{f^n}{n!}$

$=1+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}.........\frac{n!}{n!}$

$=1{+}^n{C}_1{+}^n{C}_2{+}^n{C}_3............{+}^n{C}_n$

$=2^n$