Question

If $f\left(x\right)=x^n$, then the value of $f\left(1\right)+\frac{f^1\left(1\right)}{1!}+\frac{f^2\left(1\right)}{2!}+......\frac{f^n\left(1\right)}{n!}$, where $f^r\left(x\right)$ denotes the $f^r\left(x\right)$ derivative of $r^{th}$ w.r.t.x

Answer 1

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

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

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

$f^r\left(n\right)=n(n-1)(n-2)(n-3)(n-r+1)x^{n-r}$

$=\frac{n!}{(n-r)!}{x}^{n-r}$

$S\left(1\right)=f\left(1\right)+\frac{f^1\left(1\right)}{1!}+\frac{f^2\left(1\right)}{1!}+.....\frac{f^n\left(1\right)}{n!}$

${S\left(x\right)={\sum }_{r=0}^n\frac{f^r\left(x\right)}{r!}={\sum }_{r=0}^n\frac{n!}{r!(n-r)!}{x}^{n-r}|}_{atx=1}$

[ substituting $f^r\left(x\right)$ from above]

$S\left(x\right)={\sum }_{r=0}^n{}^n{C}_r{x}^{n-r}={\sum }_{r=0}^n{}^n{C}_r{1}^r{x}^{n-r}$

this is a binomial expansion, so,

$S\left(x\right)=(1+x{)}^n$

$S\left(1\right)=2^n$ Ans.