Question

Aratati ca: (a-1)(a'n-1+a'n-2+...+a+1)=a-1​ || Dau coroana plsss

Answer 1

$\it (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+\ ...\ +a^2+a+1)=\\ \\ =a^n+a^{n-1}+a^{n-2}+\ ...\ +a^3+a^2+a-a^{n-1}-a^{n-2}-a^{n-3}-\ ...\ -a^2-a-1=\\ \\ =a^n-1$

Answer 2

Răspuns: Ai demonstrația mai jos

Explicație pas cu pas:

$\bf \big(a-1\big)\cdot \big(a^{n-1}+a^{n-2}+...+a+1\big)=$

$\bf a^{n-1+1}+a^{n-2+1}+...+a^{1+1}+a- a^{n-1}-a^{n-2}-...-a-1=$

$\bf a^{n}+\not a^{n-1}+\not a^{n-2}+...+\not a^{2}+\not a-\not a^{n-1}-\not a^{n-2}-...-\not a-1=$

$\red{\underline{\bf ~a^{n}-1~}}$