Re:[obm-l] somatorio

De:

owner-obm-l@mat.puc-rio.br

Para:

obm-l@mat.puc-rio.br

Cópia:

Data:

Sat, 27 May 2006 03:41:49 +0000 (GMT)

Assunto:

[obm-l] somatorio

> Calcule : sum(k=0->n)k^2*C(n,k)*5^k

> gab: 5n(5n+1)6^(n-2).

Usando repetidamente o fato de que k*C(n,k) = n*C(n-1,k-1), temos:

k^2*C(n,k) =

k*n*C(n-1,k-1) =

n*(k-1)*C(n-1,k-1) + n*C(n-1,k-1) =

n*(n-1)*C(n-2,k-2) + n*C(n-1,k-1).

Logo, a soma fica:

n*(n-1)*SOMA(k=0...n) C(n-2,k-2)*5^k +

+ n*SOMA(k=0...n) C(n-1,k-1)*5^k =

n*(n-1)*5^2*SOMA(j=0...n-2) C(n-2,j)*5^j +

n*5*SOMA(j=0...n-1) C(n-1,j)*5^j =

25*n*(n-1)*(1 + 5)^(n-2) + 5*n*(1 + 5)^(n-1) =

(25*n^2 - 25*n + 30*n)*6^(n-2) =

5*n*(5*n+1)*6^(n-2)

[]s,

Claudio.