Question

已知数列{f(n)}的前n项和Sn,且Sn=n^2+2n .求：若a1=f(1),an+1=f(an)(n属于N*),求

Answer 1

解的f(n)=sn-sn-1=2n+1(n>=2),f(1)=3

即f(n)=2n+1(n>=1),

an+1=f(an)=an+1=2an+1

an+1+1=2(an+1),a1=3

所以｛an+1}是等比数列.公比为2,首项为4