Question

I: lf $f\left(x\right)=(1+x{)}^n$, then the value of $f\left(0\right)+f^{\prime }\left(0\right)+\frac{1}{2!}{f}^{\prime \prime }\left(0\right)+\cdots +\frac{1}{n!}$ $f^n\left(0\right)$ is $2^n$
II: $f\left(x\right)=x^n+4$, then the value of $f\left(1\right)+f^{{}^{\prime }}\left(1\right)+\frac{1}{2!}{f}^{{}^{\prime \prime }}\left(1\right)+\cdots +\frac{1}{n!}{f}^n\left(1\right)$ is $2^n+4$.
Which of the following is correct?

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

The correct option is C Both I and II are true

$I:f\left(0\right)+f^{{}^{\prime }}\left(0\right)+\frac{1}{2!}{f}^{{}^{\prime \prime }}\left(0\right)+\cdots \frac{+}{1}n!f^n\left(0\right)$
$=1+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}+\cdots +\frac{n!}{n!}$
$={}^n{C}_0+{}^n{C}_1+{}^n{C}_2+\cdots +{}^n{C}_n$
$=2^n$
$II:f\left(1\right)+f^{\prime }\left(1\right)+\frac{f^{\prime \prime }\left(1\right)}{2!}+\cdots +\frac{1}{n!}{f}^n\left(1\right)$
$=1+4+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}+\cdots +\frac{n!}{n!}$
$=4+{}^n{C}_0+{}^n{C}_1+{}^n{C}_2\cdots +{}^n{C}_n$
$=4+2^n$
$\therefore$ Both are true.