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=2n*an,则b1+b2+…+bn的结果为

Question

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=2n*an,则b1+b2+…+bn的结果为

Answer 1

S3=3a1+3d≤9
a1+d≤3
a2≤3
a4>3,a4>a2,a4=a2+2d>a2 d>0,公差d>0,数列为递增数列.
a1>1,又a1为整数,a1至少为2,又a2≤3,a1至多为2,因此a1=2 a2=3
d=a2-a1=3-2=1
an=a1+(n-1)d=2+n-1=n+1
bn=2n×an=2n(n+1)=2n²+2n
Tn=b1+b2+...+bn
=2(1²+2²+...+n²)+2(1+2+...+n)
=2n(n+1)(2n+1)/6 +2n(n+1)/2
=[n(n+1)/3][(2n+1)+3]
=2n(n+1)(n+2)/3