Question

Sa se demonstreze relatia:

$C^{K+1 }_{2K+2} - C^{K }_{2K} = \frac{3K+1 }{K} C^{K+1 }_{2K}$

Answer 1

C2k+2^(k+1)=(2k+2)!/(k+1)!*(2k+2-k-1)!=(k+1)!*(k+2)..(2k+1)(2k+2)/(k+1)!*(k+1)!=
(k+2)*...*(2k+1)(2k+2)/(k+1)! Relatia (A
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C2k^k=(2k)!/k!*(2k-k)!=k!*(k+1)(k+2)...(2k)/k!*(2k-k)!=k!*(k+1)(k+2)...(2k)!/k!*k!=
(k+1)(k+2)...(2k)!/k!
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(3k+1)/k*C2k^(k+1)=(3k+1)/k*(2k)!/(k+1)!(2k-k-1)=(3k+1)/k*(2k)!/(k+1)!*(k-1)!=
(3k+1)/k*(k+1)!*(k+2)...(2k)/(k+1)!*(k-1)!=(3k+1)/k*(k+2)...(2k)/(k-1)!=
(3k+1)/k*(k+2)...(2k)/(k-1) !     (C
Inlocuiesti  valorile  (A  si  (B    in  relatia  data  si  obtii (C
(k+2)...(2k+1)(2k+2)/(k+1)!-(k+1)(k+2)...(2k)/k!=aduci  la  acelasi  numitor
[(k+2)...(2k+1)(2k+2)-(k+1)²]/(k+1)!=(k+2)...(2k)*[(3k²+4k+1)]/(k+1)!=
(k+2)...(2k)*(k+1)(3k+1)/(k+1)!=(k+2)...(2k)*(3k+1)/k*(k-1)!=(C