Question

folosind metoda inductiei matematice sa se demonstreze ca pentru orice nr natural sunt adevarate egalitatile 1*3+2*4+3*5+...+ n( n+2) = n(n+1)(2n+1) / 6

Answer 1

P(n):1*3+2*4+3*5+....+n(n+2)=n(n+1)(2n+1)/6

Etapa I(de verificare)
P(1): 1*3=1(1+1)(2*1+1)/6==>3=2*3/6==>3=6/6==>6/6=6/6==>1=1 se adice la acelasi numitor

Etapa aIIa(de demonstrare)
Se presupune P(k). (A) ,k>1
Se demonstreaza P(k+1) (A) ,unde P(k)-->P(k+1)

P(k):1*3+2*3+3*5+....+k(k+2)=k(k+1)(2k+1)/6
P(k+1):1*3+2*3+3*5+....+k(k+2)+(k+1)(k+3)=(k+1)(k+2)(k+3)/6
de acum se inlocueste k(k+1)(2k+1)/6+(k+1)(k+3)=(k+1)(k+2)(k+3)/6==> si acum faci calculele

Sper ca te-am ajutat