数列{an}的前n项和为Sn,且a1=1,an+1=13Sn(n=1,2,3,.)求(1)a2,a3,a5的直及数列{an}的通项式：

题目

数列{an}的前n项和为Sn,且a1=1,an+1=13Sn(n=1,2,3,.)求(1)a2,a3,a5的直及数列{an}的通项式：
(2)a2+a4+a6+.+a2n的值,

答案

a1= 1,s1 = a1 = 1
a2 = (1/3)*s1 = 1/3
s2 = a1+a2 = 1+1/3 = 4/3
a3 = (1/3)*s2 = 4/9
s3 = a1+a2+a3 = 4/3+4/9 = 16/9
a4 = (1/3)*s3 = 16/27
s4 = s3+a4 = 16/9 + 16/27 = 64/27
a5 = (1/3)*s4 = 64/81
有了前5项,数组的通项式可以得到了 an = 4^(n-2)/3^(n-1) [其中^是指数的意思] n从2开始.
有了通项式第二问就可以自己做了