Question

{an}是各项为正的等比数列,bn是等差数列,且a1=b1=1,a3+b5=13,a5+b3=21,Sn为an前n项和,求{Sn×bn}前n项和tn

Answer 1

令an的公比为q，bn的公差为d
a3+b5=q^2+1+4d=13,
a5+b3=q^4+1+2d=21
∵{an}各项为正,q>0
∴d=2,q=2
Sn=a1(1-q^n)/(1-q)=2^n-1
bn=a1+(n-1)d=2n-1
Sn*bn=(2n-1)(2^n-1)
Tn=(2^1-1)+3(2^2-1)+5(2^3-1)+...+(2n-1)(2^n-1)
=2^1+3*2^2+5*2^3+...+(2n-1)*2^n-n^2
2Tn=2^2+3*2^3+5*2^4+...+(2n-1)*2^(n+1)-2n^2
2Tn-Tn=2^2+3*2^3+5*2^4+...+(2n-1)*2^(n+1)-2n^2-[2^1+3*2^2+5*2^3+...+(2n-1)*2^n-n^2]
Tn=(2n-1)*2^(n+1)-2^2-2(2^3+2^4+2^5+...+2^n)-n^2
=(2n-3)*2^(n+1)-n^2+12