Question

Ajutor cu tema de mate va rog
Clasa a 9 a​

Answer 1

P(n):1-2-3+3-4-5+...+(2n-1)2n(2n+1)=n(n+1)

(2n²+2n-1)

Etapa 1) Verificam daca P(1)(A) P(1):1-2-3-1(1+1)(2*12+2*1-1) (A) deoarece 6=6.

Etapa2)

Presupunem ca P(n)(A) si aratam ca P(n+1)(A). P(n+1):1-2-3+3.4.5+...+(2n-1)2n(2n+1)+(2n+1)(2n+2) (2n+3)=(n+1)(n+2)[2(n+1)²+2(n+1)-1)]=(n+1)(n+2) (2n²+4n+2+2n+2-1)=(n+1)(n+2)(2n²+6n+3)

1-2-3+3.4.5+...+(2n-1)2n(2n+1)+(2n+1)(2n+2)(2n+3)=

=n(n+1)(2n²+2n-1)+(2n+1)(2n+2)(2n+3)=

=n(n+1)(2n²+2n-1)+(2n+1)2(n+1)(2n+3)= =(n+1)(2n³+2n²-n+8n²+12n+4n+6)=

=(n+1)(2n³+10n²+15n+6)= =(n+1)(n+2)(2n²+6n+3)

Sper ca te-am ajutat. Coroană?