Question

If $f\left(x\right)=x^n$ then value of $f\left(1\right)-\frac{f^{\prime }\left(1\right)}{1!}+\frac{f^{\prime \prime }\left(1\right)}{2!}-\frac{f^{\prime \prime \prime }\left(1\right)}{3!}+....+\frac{{(-1)}^n{f}^n\left(1\right)}{n!}$ is

• A. $2^{n-1}$
• B. $0$
• C. $1$
• D. $2^n$

Answer 1

The correct option is B $0$

$f\left(x\right)-\frac{f^{\prime }\left(x\right)}{1!}+\frac{f^{\prime \prime }\left(x\right)}{2!}-....+\frac{(-1{)}^n{f}^n(x)}{n!}$
$=x^n-\frac{n{x}^{n-1}}{1!}+\frac{n(n-1)x^{n-2}}{2!}+...+\frac{(-1{)}^{n-1}x}{(n-1)!}+\frac{(-1{)}^n}{n!}$
$={}^n{C}_0{x}^n-{}^n{C}_1{x}^{n-1}+{}^n{C}_2{x}^{n-2}+...+(-1{)}^n{}^n{C}_n{x}^0$
$=(x-1{)}^n$
Now substituting $x=1$, we get
The sum of the series as $0$.