{an]前n项和为Sn,2Sn=a(n+1)—2的n+1次方+1,且a1,a2 +5 a3成等差数列（1）求a1 (2)求数列an通项公式

a1,a2+5,a3成等差数列a1+a3 = 2(a2+5) (1)2Sn=a(n+1)-2^(n+1) +1for n>=2an = Sn -S(n-1)2an =a(n+1)-an -2^na(n+1)= 3an +2^na(n+1)+ 2^(n+1) = 3[ an + 2^n]{an + 2^n } 是等比数列,q=3an + 2^n = 3^(n-1) .( a1 +...