Question

P(n):1•2•3+2•3•4+....+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4​

Answer 1

P(n):1*2*3+2*3*4+...+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4,∀n≥1

| Verificarea:n=1=>P(1):1(1+1)(1+2)=1(1+1)(1+2)(1+3)/4<=>2*3=2*3*4/4<=>6=6 (A)

|| Demonstratia:P(k)->P(k+1),∀k≥1

fie v(P(k))=1

P(k):1*2*3+2*3*4+...+k(k+1)(k+2)=k(k+1)(k+2)(k+3)/4,∀k≥1

P(k+1):1*2*3+2*3*4+...+(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)(k+4)/4,∀k≥1

_____________________________________________\\__

1*2*3+2*3*4+...+k(k+1)(k+2)+(k+1)(k+2)(k+3)=k(k+1)(k+2)(k+3)/4+(k+1)(k+2)(k+3)=[k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)]/4=(k+1)(k+2)(k+3)(k+4)/4 (A),∀k≥1=>v(P(k+1))=1=>P(k)->P(k+1),∀k≥1=>v(P(n))=1,∀n≥1