急!数列{an}中,若Tn=a1a2a3…=n(n+1)(n+2) (n>=1),则此数列的通项公式an=?
问题描述:
急!数列{an}中,若Tn=a1a2a3…=n(n+1)(n+2) (n>=1),则此数列的通项公式an=?
答
Tn=a1a2a3…an=n(n+1)(n+2)
当n=1时,a1=T1=1*2*3=6
当n≥2时,anTn/T(n-1)=[n(n+1)(n+2)]/[(n-1)n(n+1)]=(n+2)/(n-1)
∴此数列的通项公式
an={ 6,(n=1)
{ (n+2)/(n-1) ,(n≥2)