If $f(x)=x^n$, then the value of $f(1)-\frac{f^{\prime}(1)}{1 !}+\frac{f^{\prime \prime}(1)}{2 !}-\frac{f^{\prime \prime \prime}(1)}{3 !}+\frac{f^{i v}(1)}{4 !} ....+\frac{(-1)^n f^n(1)}{n !}$, is
Answer & explanation
Correct answer: option 4
We have,
$f(x)=x^n$
$f^r(x)=n(n-1)(n-2) .....(n-(r-1)) x^{n-r}$
$\Rightarrow f^r(x)=\frac{n !}{(n-r) !} x^{n-r} \Rightarrow f^r(1)=\frac{n !}{(n-r) !}$
∴ $f(1)-\frac{f^{\prime}(1)}{1 !}+\frac{f^{\prime \prime}(1)}{2 !}-\frac{f^{\prime \prime \prime}(1)}{3 !}+...+\frac{(-1)^n f^n(1)}{n !}$
$= \sum\limits_{r=0}^n(-1)^r \frac{f^r(1)}{r !}$, where $f^0(1)=f(1)$
$= \sum\limits_{r=0}^n(-1)^r \frac{n !}{(n-r) ! r !}=\sum\limits_{r=0}^n(-1)^r~{ }^n C_r=0$