←Vrati se na zadatke1. ZadatakOdredite prirodan broj n≥2n\ge2n≥2 tako da vrijedi jednakost (n−1)2n(n+1)!(n+2)!=(n2).\frac{(n-1)^2n(n+1)!}{(n+2)!}=\binom n2.(n+2)!(n−1)2n(n+1)!=(2n).Prikaži rješenjeRješenje(n−1)(n−1)n(n+1)!(n+2)!=n!2!(n−2)!\frac{(n-1)(n-1)n(n+1)!}{(n+2)!}=\frac{n!}{2!(n-2)!}(n+2)!(n−1)(n−1)n(n+1)!=2!(n−2)!n! n−1n+2=12\frac{n-1}{n+2}=\frac12n+2n−1=21 2n−2=n+22n-2=n+22n−2=n+2 n=4n=4n=4Pomakni formulu lijevo ili desno.