Răspuns:
n ∈ N - {numerele de forma 3k}
Explicație pas cu pas:
²⁾1/3 + ³⁾n/2 + n²/6 = (n²+3n+2)/6 =
= (n²+n+2n+2)/6 = [n(n+1)+2(n+1)]/6 =
= (n+1)(n+2)/6
n = 1 => 2·3/6 = 1 ∈ N
n = 2 => 3·4/6 = 2 ∈ N
n = 4 => 5·6/6 = 5 ∈ N
n = 5 => 6·7/6 = 7 ∈ N
n = 7 => 8·9/6 = 12 ∈ N
n = 8 => 9·10/6 = 15 ∈ N
n = 10 => 11·12/6 = 22 ∈ N
n = 11 => 12·13/6 = 26 ∈ N
n = 13 => 14·15/6 = 35 ∈ N
n = 14 => 15·16/6 = 40 ∈ N
n = 16 => 17·18/6 = 51 ∈ N
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Solutie : n ∈ N - {numerele de forma 3k}
(toate numerele naturale in afara de numerele de forma 3k)
