等差数列a1和d为整数 a1大于1 a4大于3 S3小于等于9 bn=2的n次方*an 求b1+b2+ ……+bn
3个回答

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

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