Cevap: C)6
Açıklama:
[tex]\frac{(n+1)! + (n-1)! }{n^3 -1} = 24\\\\\frac{(n+1)(n)*(n-1)! + (n-1)!}{(n-1)(n^2+n+1)} = 24\\ \\\frac{(n-1)! * [(n+1)*n+1]}{(n-1)(n^2+n+1)} = 24\\\\\frac{(n-1)! * (n^2+n+1)}{(n-1)(n^2+n+1)} = 24\\\\\frac{(n-1)*(n-2)!}{n-1} = 24\\ \\(n-2)! = 4!\\\\n = 6[/tex]
