Răspuns :
Răspuns:
Explicație pas cu pas:
5Cₙ³ = Cₙ₊₂⁴ ;
conditii : n ≥ 3 ; n+2 ≥4 => n ≥ 2
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5· n! /[ 3!·(n-3)!] = (n+2)! / [4!·(n+2-4)!] =>
5·n·(n-1)·(n-2)·(n-3)! / [3·2·(n-3)!] = (n+2)(n+1)(n)(n-1)(n-2)! /[4·3·2·(n-2)!] =>
5·n·(n-1)·(n-2)/6 = (n+2)(n+1)(n)(n-1)/24 =>
20(n-2) = (n+1)(n+2) =>
20n-40 = n²+3n+2 => n²-17n +42 = 0
n₁,₂ = [17±√(289-168)]/2
n₁,₂ = (17±11)/2
n₁ = (17-11)/2 = 6/2 = 3
n₂ = (17+11)/2 = 28/2 = 14
Solutii : n₁ = 3 ; n₂ = 14