Question Number 191582 by Matica last updated on 26/Apr/23
$$\:\:{f}\left({x}\right)={x}^{{n}} \:.\:\:{find}\: \\ $$$$\:\:\:{A}={f}\left(\mathrm{1}\right)+\frac{{f}^{'} \left(\mathrm{1}\right)}{\mathrm{1}}+\frac{{f}^{''} \left(\mathrm{1}\right)}{\mathrm{2}}+\frac{{f}^{'''} \left(\mathrm{1}\right)}{\mathrm{3}}+…+\frac{{f}^{\left({n}\right)} \left(\mathrm{1}\right)}{{n}} \\ $$
Answered by aleks041103 last updated on 27/Apr/23
$${f}^{\left({k}\right)} \left({x}\right)=\left({x}^{{n}} \right)^{\left({k}\right)} =\begin{cases}{\frac{{n}!}{\left({n}−{k}\right)!}{x}^{{n}−{k}} ,\:{k}=\mathrm{0},\mathrm{1},…,{n}}\\{\mathrm{0},{k}>{n}}\end{cases} \\ $$$$\Rightarrow{A}=\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{n}!}{{k}\left({n}−{k}\right)!} \\ $$