Question Number 28620 by abdo imad last updated on 28/Jan/18
$${calculate}\:\:\sum_{{n}={p}} ^{+\infty} \:\:\:{C}_{{n}\:} ^{{p}} {x}^{{n}} . \\ $$
Commented by Tinkutara last updated on 28/Jan/18
$${How}\:{can}\:{n}>{p}\:{becasuce}\:{it}\:{is}\:{used}\:{in} \\ $$$$\:^{{p}} {C}_{{n}} ? \\ $$
Commented by abdo imad last updated on 28/Jan/18
$$\sum_{{n}={p}} ^{\propto} \:{C}_{{n}} ^{{p}} \:{x}^{{n}} =\:\:{C}_{{p}} ^{{p}} \:\:{x}^{{p}} \:\:+\:{C}_{{p}+\mathrm{1}} ^{{p}} \:{x}^{{p}+\mathrm{1}} \:+… \\ $$
Commented by Tinkutara last updated on 28/Jan/18
$${But}\:{what}\:{is}\:^{{p}} {C}_{{p}+\mathrm{1}} ?\:{Is}\:{it}\:{combination}? \\ $$$${Because}\:{if}\:{we}\:{write}\:^{{n}} {C}_{{r}} \:{it}\:{means} \\ $$$${r}\leqslant{n}. \\ $$
Commented by abdo imad last updated on 28/Jan/18
$${for}\:{us}\:{we}\:{use}\:\:{C}_{{n}} ^{{p}} \:=\:\frac{{n}!}{{p}!\left({n}−{p}\right)!}\:{for}\:{p}\leqslant{n}\:\:{it}\:{s}\:{only}\:{a}\:{notation} \\ $$$${of}\:{combination}.. \\ $$