Question

等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}为等比数列,b1=2,且b2S2=32,b3S3=12

Answer 1

(1) b2*S2=(b1*q)*(a1+a1+d)=2q*(6+d)=32 => q(6+d)=16

b3*S3=(b1*q^2)*(3a1+3d)=2q^2*(9+3d)=120 => q^2*(3+d)=20

联立解得：d=2 ,q=2（因为等差数列｛an｝的各项均为正数则d>0,故d= -6/5 ,q=10/3舍掉）

则：(1) an=3+2(n-1)=2n+1

bn=2*2^(n-1)=2^n

(2) an*bn=(2n+1)*2^n

则Tn=3*2+5*2^2+.+(2n+1)*2^n

(1/2)Tn=3+5*2+.+(2n+1)*2^(n-1)

(1/2)Tn-Tn=3+2*2+2*2^2+.+2*2^(n-1)-(2n+1)*2^n

(-1/2)Tn=3+2*2*[1-2^(n-1)]/(1-2)-(2n+1)*2^n

=3+2^(n+1)-4-(2n+1)*2^n

Tn=2-2*2^(n+1)+(2n+1)*2^(n+1)

=2+(2n-1)*2^(n+1)