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cum pot sa rezolv [tex] C^{n+1}_{n+2} = 2 [/tex]

Răspuns :

(n+2)!/(n+1)!(n+2-n-1)=2
(n+2)!/(n+1)!1!=2
(n+2)(n+1)n/(n+1)n=2
(n+2)/1=2
n+2=2
n=0

Folosim formula combinărilor complementare:

[tex]\it C^k_n =C^{n-k}_n[/tex]

[tex]\it C^{n+1}_{n+2} =C^{n+2-n-1}_{n+2} =C^1_{n+2} =n+2[/tex]

Ecuatia devine:

n+2 = 2 ⇒ n = 0 .