Question

If $f(x)=x^{n},$ then the value of $f(1)+\frac{f^{1}(1)}{1}+\frac{f^{2}(1)}{2 !}+\cdots+\frac{f^{n}(1)}{n !},$ where $f^{r}(x)$ denotes the rth order derivative of $f(x)$ with respect to $x$, is

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

Answer 1

Answer: (B)

We have $f(x)=x^{n} .$ So,
$f^{1}(x)=n x^{n-1} \Rightarrow f^{\prime}(1)=n$
$f^{2}(x)=n(n-1) x^{n-2} \Rightarrow f^{2}(1)=n(n-1)$
$f^{3}(x)=n(n-1)(n-2) x^{n-3} \Rightarrow f^{3}(1)=n(n-1)(n-2)$
$ f^{n}(x)=n(n-1)(n-3) \ldots 1 \Rightarrow f^{n}(1)=n(n-1)(n-3) \ldots 1 $
$\Rightarrow f(1)+\frac{f^{1}(1)}{1}+\frac{f^{2}(1)}{2 !}+\cdots+\frac{f^{n}(1)}{n !} $
$=1+\frac{n}{1}+\frac{n(n-1)}{2 !}+\frac{n(n-1)(n-2)}{3 !}+\cdots+\frac{n(n-1)(n-2) \ldots 1}{n !}$
$={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\cdots+{ }^{n} C_{n}$
$=2^{n}$