Suma combinatorica

[Bianca1215]

Ma puteti ajuta cu demonstrarea acestor sume? Multumesc!

[Integrator]

Ma puteti ajuta cu demonstrarea acestor sume? Multumesc!

Oidee:
Încercati să demonstrati prin inductie matematică.

[DD]

2)Termenii sumei se pot scrie si k*∑_(i=1)^k▒C_k^i =k*(2^k-1)si suma ceruta deType equation here.vine
∑_(k=1)^n▒(k*2^k-k) ,unde;
∑_(k=1)^n▒〖(k*2^k )=〗 2^n+………………………………………………………………………………=2^n*(2-1)
2^n 〖+2〗^(n-1)+………………………………………………………………………=2^(n-1 )*(2^2-1)
………………………….+2^(n )+2^(n-1)+2^(n-2)+⋯……………………………………..=2^(n-2)*(2^3-1)
………………………….+2^n+2^(n-1)+2^(n-2)+2^(n-3)+……………………………………………=2^(n-3)(〖 2〗^4-1)
………………………………………………………………………………………………………………………………………………..
…………………………+2^n+2^(n-1)+2^(n-2)+2^(n-3)+2^(n-4)+⋯………+2^2+⋯=2^2*(2^(n-1)-1)
……………………+.2^n+2^(n-1)+2^(n-2)+⋯………………………,,,+2^2+2^1=2*(2^n-1)
∑_(k=1)^n▒〖(k*2^k )=〗 n*2^(n+1)-2^(n+1)+2=2^(n+1)*(n—1)+2
∑_(k=1)^n▒〖(k)=n*〗(1+n)/2
∑_(k=1)^n▒〖(k*2^k-k)=〗 (n-1)*2^(n+1)+2-n*(n+1)/2

[DD]

1)Se cere calculeze suma;∑_(k=0)^n▒(1/(k+1)(k+2) )(C_n^k )
Expresia; (1/(k+1)(k+2) )(C_n^k )=(1/(n+1)(n+2) )(C_(n+2)^(k+2) )si ∑_(k=0)^n▒(1/(n+1)(n+2) )(C_(n+2)^(k+2) ) =
(1/(n+1)(n+2) )*(±1±(n+2)+C_(n+2)^2+C_(n+2)^3+……………………C_(n+2)^(n+2))= (1/(n+1)(n+2) )*(2^(n+2)-(n+3))