Question

数列{an}的前n项和为Sn，且a1=1，an+1=1\3Sn(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，要过程谢谢！

Answer 1

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