数列{an}的前n项和为SN,且a1=1.an+1=1/3sn,n=1,2,3...

题目

数列{an}的前n项和为SN,且a1=1.an+1=1/3sn,n=1,2,3...
求(1)数列{an}通项公式 （2）求a1+a2+a3+..+a2n-1的值

答案

a(n+1) = sn/3,
an = s(n-1)/3,
两式相减得
a(n+1) - an = [sn - s(n-1)]/3 = an/3
a(n+1) = 4an/3
a(n+1) /an=4/3
{an} 是首项为1,公比为4/3的等比数列,
an=a1q^(n-1)
an = (4/3)^(n-1),n = 1,2,...
a1+a2+a3+..+a2n-1
=[1-(4/3)^(2n-1)]/(1-4/3)
=3[(4/3)^(2n-1)-1]
=3*(4/3)^(2n-1)-3