(1)设等差数列{an}的公差为d,
则依题意可知d>0由a2+a7=16,
得2a1+7d=16①
由a3a6=55,得(a1+2d)(a1+5d)=55②
由①②联立方程求得
得d=2,a1=1或d=-2,a1= 20 7 (排除)
∴an=1+(n-1)•2=2n-1
(2)令cn= bn 2n ,则有an=c1+c2+…+cn
an+1=c1+c2+…+cn+1
两式相减得
an+1-an=cn+1,由(1)得a1=1,an+1-an=2
∴cn+1=2,即cn=2(n≥2),
即当n≥2时,
bn=2n+1,又当n=1时,b1=2a1=2
∴bn=
2,(n=1) 2n+1,(n≥2)
于是Sn=b1+b2+b3+…+bn=2+23+24+…2n+1=2n+2-6,n≥2,
Sn=
2n=1 2n+2-6n≥2 .
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