Question

arata ca:(a-1)(a la n-1+a la n-2+...+a+1)=a la n-1

Answer 1

$S=a^{n-1}+a^{n-2}+\ldots+a+1.\\aS=a^n+a^{n-1}+\ldots+a^2+a.\\aS-S=(a-1)S=a^n+a^{n-1}+\ldots+a^2+a - (a^{n-1}+a^{n-2}+\ldots+a+1)=a^n-1.\\Deci\ (a-1)S=a^n-1,\ ceea\ ce\ trebuia\ demonstrat.$

Simplu, nu ? :-).

Green eyes.