Odpowiedzi

2009-11-22T16:52:45+01:00
A)
(n-1)!/(n+1)!=(n-1)!/[(n-1)!*n*(n+1)]=1/[n*(n+1)]

b)
(n+3)!/n!= n!(n+1)(n+2)(n+3)/n! = (n+1)(n+2)(n+3)
2009-11-22T16:54:26+01:00
A)(n-1)!/(n+1)!=-1
b) (n+3)!/n!= n+2*n+3