Principio di induzione

E' lungo e laborioso.

Ponendo n + 1 al posto di n scrivi

(1 + 1/(n+1))^n = S_k:0->n  C(n,k) 1/(n+1)^k

Pertanto

( 1 + 1/(n+1))^(n+1) = (1 + 1/(n+1)) * (1 + 1/(n+1))^n =

= (1 + 1/(n+1)) * S_k:0->n  C(n,k) 1/(n+1)^k =

= S_k:0->n  C(n,k) 1/(n+1)^k + S_k:0->n  C(n,k) 1/(n+1)^(k+1) =

= C(n,0) 1/(n+1)^0 +

+ C(n,0) 1/(n+1)^1 + C(n, 1) 1/(n + 1)^1 +

+ C(n,1) 1/(n+1)^2 + C(n,2) 1/(n+1)^2 +

.............................

+ C(n,n-1)  1/(n+1)^n + C(n,n) 1/(n+1)^n +

+ C(n,n) 1/(n+1)^(n+1)

che significa anche

C(n+1,0) 1/(n+1)^0 + S_k:1->n   [C(n, k-1) + C(n, k)] 1/(n+1)^k +

+ C(n+1, n+1) 1/(n+1)^(n+1)

Ora C(n, k-1) + C(n, k) = n!/[(k-1)!(n-k+1)!] + n!/[k!(n-k)!] =

= [n!(k + n-k+1)/[k!(n-k+1)!] = (n!(n+1))/(k!(n+1-k)!) = (n+1)!/(k!(n+1-k)!) =

= C(n+1, k)

Pertanto la somma precedente si riscrive

C(n+1,0) 1/(n+1)^0 + S_k:1->n   C(n+1, k)  1/(n+1)^k +

+ C(n+1, n+1) 1/(n+1)^(n+1) =

= S_k:0->n+1   C(n+1, k) 1/(n+1)^k

che é proprio la versione della formula riportata per il successivo n+1.