{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

令an的公比为q,bn的公差为da3+b5=q^2+1+4d=13,a5+b3=q^4+1+2d=21∵{an}各项为正,q>0∴d=2,q=2Sn=a1(1-q^n)/(1-q)=2^n-1bn=a1+(n-1)d=2n-1Sn*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*...