Question

Inductie matematica:
1·2·3+3·4·5+...+(2n-1)2n(2n+1)=n(n+1)(2n²+2n-1)

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)