已知等差数列{an}的首项a1及公差d都是整数,前n项和为Sn,若a1>1,a4>3,S3≤9,设bn=2的n次方乘以a

设公差为d.
S3=3a1+3d=3(a1+d)=3a2≤9
a2≤3
又a4>3 a4-a2=2d>0 d>0,又d为整数,d≥1
a1为整数,又a1>1,因此a1≥2
a2=a1+d≥2+1=3,又a2≤3,因此只有a2=3 a1=2 d=1
an=a1+(n-1)d=2+n-1=n+1
bn=2ⁿ×an
Tn=b1+b2+...+bn=2×2+3×2²+4×2³+...+(n+1)×2ⁿ
2Tn=2×2²+3×2³+...+n×2ⁿ+(n+1)×2^(n+1)
Tn-2Tn=-Tn=2×2+2²+2³+...+2ⁿ -(n+1)×2^(n+1)
=1+1+2+2²+2³+...+2ⁿ -(n+1)×2^(n+1)
=1+1×[2^(n+1) -1]/(2-1) -(n+1)×2^(n+1)
=-n×2^(n+1)
b1+b2+...+bn=Tn=n×2^(n+1)