高一数学. 已知等差数列{an}的前n项和胃Sn,公差d≠0,且S3+S5=50,a1,a4,a13成等比数列．

a(n)=a+(n-1)d,
s(n)=na+n(n-1)d/2.
[a(4)]^2=a(1)a(13)=[a+3d]^2=a[a+12d],
a^2 + 6ad + 9d^2 = a^2 + 12ad,
0 = 9d^2 - 6ad = 3d[3d-2a],a = 3d/2.
a(n)=3d/2 + (n-1)d.
s(n)=3nd/2 + n(n-1)d/2.
50=s(3)+s(5)=3*3d/2 + 3d + 3*5d/2 + 10d=25d,d=2.
a(n)=3+2(n-1)=2n+1.
b(n)=a(2n)=4n+1
t(n)=b(1)+b(2)+...+b(n)=2n(n+1)+n
若是
b(n)=a[2^n]=2*2^n + 1 = 4*2^(n-1)+1,
t(n)=4[2^n - 1]/(2-1) + n = 4[2^n - 1] + n = 2^(n+2) - 4 + n

（1）S3+S5=3a1+3d+5a1+10d=50 (等差数列和公式Sn=na1+n(n-1)d/2)
推出 8a1+13d=50 ①
a1 ,a4, a13成等比数列：(a4)^2=a1*a3
(等差数列通项公式an=a1+(n-1)d)
a4=a1+3d; a13=a1+12d
推出 3d=2a1 ②
将②带入①中，得...