Question

să se verifice daca:

1!•0+2!•1²+3!•2²+...+n!•(n-1)² = (n+1)!(n-2)+2

Mulțumesc!​

Answer 1

Răspuns:

Explicație pas cu pas:

Prin inductie:

P(n) = 1!•0+2!•1²+3!•2²+...+n!•(n-1)² = (n+1)!(n-2)+2

P(3) = 0+2!*1 +3!*2^2 = 2 + 1*2*3*4 = 26 = (3+1)!*(3-2) +2

P(3) se verifica, presup.adevrata P(n),

sa aratam ca P(n+1) = (n+2)!(n-1) +2

P(n+1) =1!•0+2!•1²+3!•2²+...+n!•(n-1)²  + (n+1)!*n^2 =

p(n) +(n+1)!*n^2  = (n+1)!(n-2)+2 +(n+1)!*n^2 =

(n+1)!(n^2 +n -2) +2 = (n+1)!(n+2)(n-1) +2  

(n+2)!(n-1) +2