设{an}公差为d
a1,a4,a3成等比数列
a4²=a1·a3
(a5-d)²=(a5-4d)(a5-2d)
4da5-7d²=0
a5=7代入,整理,得d²-4d=0
d(d-4)=0
d=0(与已知矛盾,舍去)或d=4
a1=a5-4d=7-4×4=-9
an=a1+(n-1)d=-9+4(n-1)=4n-13
bn=a(2ⁿ)=4·2ⁿ-13=2^(n+2) -13
Sn=b1+b2+...+bn
=[2³+2⁴+...+2^(n+2)] -13n
=8×(2ⁿ-1)/(2-1) -13n
=2^(n+3) -13n -8