{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)
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a1,a2+5,a3成等差数列
a1+a3 = 2(a2+5) (1)
2Sn=a(n+1)-2^(n+1) +1
for n>=2
an = Sn -S(n-1)
2an =a(n+1)-an -2^n
a(n+1)= 3an +2^n
a(n+1)+ 2^(n+1) = 3[ an + 2^n]
{an + 2^n } 是等比数列,q=3
an + 2^n = 3^(n-1) .( a1 + 2)
a1 = (an + 2^n)/3^(n-1) -2 (2)
an + 2^n = 3^(n-2) .( a2 + 4)
a2 = (an + 2^n)/3^(n-2) -4 (3)
an + 2^n = 3^(n-3) .( a3 + 8)
a3 = (an + 2^n)/3^(n-3) -8 (4)
sub (2),(3),(4) into (1)
a1+a3 = 2(a2+5)
(an + 2^n)/3^(n-1) +(an + 2^n)/3^(n-3) -10 = 2[(an + 2^n)/3^(n-2) +9]
(an + 2^n)/3^(n-1) +(an + 2^n)/3^(n-3) = 2(an + 2^n)/3^(n-2) +28
(an + 2^n) +9(an + 2^n) = 6(an + 2^n) +28.3^(n-1)
4an = 28.3^(n-1) -4.2^n
an = (7/3).3^n - 2^n
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