Question

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

• A. $2^n$
• B. $0$
• C. $2^{n-1}$
• D. none of these

Answer 1

The correct option is B $0$

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

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

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

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

Now,

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

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

$={}^n{C}_0-{}^n{C}_1+{}^n{C}_2-{}^n{C}_3+...+{}^n{C}_n$

We know that the alternate sum and difference of binomial coefficients is $0$.

Hence, the value of the given expression is $0$.