Question Number 31229 by abdo imad last updated on 03/Mar/18
$${simplify}\:{A}_{{n}} =\:{C}_{{n}} ^{\mathrm{1}} \:\:+\mathrm{2}\:{C}_{{n}} ^{\mathrm{2}} \:+\mathrm{3}\:{C}_{{n}} ^{\mathrm{3}} \:+...\:+\:{n}\:{C}_{{n}} ^{{n}} \:. \\ $$
Commented by abdo imad last updated on 05/Mar/18
$${let}\:{introduce}\:{the}\:{polynomial}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \\ $$$${we}\:{have}\:{p}\left({x}\right)=\:\left({x}+\mathrm{1}\right)^{{n}} \:\Rightarrow{p}^{'} \left({x}\right)={n}\left({x}+\mathrm{1}\right)^{{n}−\mathrm{1}} \:{from}\: \\ $$$${another}\:{side}\:{p}^{'} \left({x}\right)=\:\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:{C}_{{n}} ^{{k}} \:{x}^{{k}−\mathrm{1}} \:\:{and}\:{we}\:{have} \\ $$$${A}_{{n}} ={p}^{'} \left(\mathrm{1}\right)=\:{n}\:\mathrm{2}^{{n}−\mathrm{1}} \:. \\ $$$$ \\ $$