Aratati ca: (a-1)(a'n-1+a'n-2+...+a+1)=a-1 || Dau coroana plsss
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[tex]\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[/tex]
Răspuns: Ai demonstrația mai jos
Explicație pas cu pas:
[tex]\bf \big(a-1\big)\cdot \big(a^{n-1}+a^{n-2}+...+a+1\big)=[/tex]
[tex]\bf a^{n-1+1}+a^{n-2+1}+...+a^{1+1}+a- a^{n-1}-a^{n-2}-...-a-1=[/tex]
[tex]\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=[/tex]
[tex]\red{\underline{\bf ~a^{n}-1~}}[/tex]