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2014-05-04T17:01:37+03:00
(n-1)!n(n+1)=5(n-1)!n+12(n-1)!⇒impartim toata ec cu (n-1)! si obtinem n²+n=5n+12⇒
n²-4n-12=0 Δ=64, n1=6, n2=-2⇒ S=6∈N
(n-1)!*n*(n+1)(n+2)(n+3)/(n-1)!=14(n-2)!(n-1)n(n+1)/(n-2)!⇒n(n+1)(n+2)(n+3)=14(n-1)n(n+1) ⇒n(n+1)(n²+5n+6-14n+14)=n(n+1)(n²-9n+20)=0⇒S=(0,4, 5), iar n=-1 nu apartine lui N
n!(n+1)(n+2)(n+3)(n+4)/n!=360⇒n1=2, 
2 5 2