已知数列{An},{Bn}都是无穷等差数列,其中a1=3,b1=2,b2是a2与a3的等差中项,且limAn/Bn=1/2,求lim(1/(a1b1)+1/(a2b2)+……+1/(anbn)的值.

已知数列{An},{Bn}都是无穷等差数列,其中a1=3,b1=2,b2是a2与a3的等差中项,且limAn/Bn=1/2,求lim(1/(a1b1)+1/(a2b2)+……+1/(anbn)的值.
已知数列{An},{Bn}都是无穷等差数列,其中a1=3,b1=2,b2是a2与a3的等差中项,且limAn/Bn=1/2,求lim(1/(a1b1)+1/(a2b2)+……+1/(anbn)的值.

已知数列{An},{Bn}都是无穷等差数列,其中a1=3,b1=2,b2是a2与a3的等差中项,且limAn/Bn=1/2,求lim(1/(a1b1)+1/(a2b2)+……+1/(anbn)的值.
设{An}公差为d,{Bn}公差为d'
An=[2a1+(n-1)d]*n/2
Bn=[2b1+(n-1)d']*n/2
lim An/Bn=lim [6+(n-1)d]/[4+(n-1)d']=lim [6/(n-1)+d]/[4/(n-1)+d']=d/d'
因为limAn/Bn=1/2,所以d/d'=1/2, d'=2d (1)
因为b2是a2与a3的等差中项,所以a1+d+a1+2d=2(b1+d'),
2*3+3d=2*2+2d' 2d'=3d+2 (2)
联立(1)(2) 解得d=2,d'=4
所以an=a1+(n-1)*2=2n+1, bn=b1+(n-1)*4=4n-2=2(2n-1)
所以lim[1/(a1b1)+1/(a2b2)+……+1/(anbn)]
=lim(1/2)[1/1*3+1/3*5+.+1/(2n-1)(2n+1)]
=lim(1/2)*{(1/2)(1-1/3)+(1/2)(1/3-1/5)+(1/2)[1/(2n-1)-1/(2n+1)]}
=lim(1/2)*(1/2)[1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1)]
=lim(1/4)*[1-1/(2n+1)]
=lim(1/4)*[2n/(2n+1)]
=1/4

设An公差为d，Bn公差为D
An=(a1+an)n/2
Bn=(b1+bn)n/2
An/Bn=(a1+an)/(b1+bn)=[6+(n-1)d]/[4+(n-1)D]
因为limAn/Bn=1/2，所以的d/D=1/2,D=2d
因为b2是a2与a3的等差中项，所以a1+d+a1+2d=2(b1+D),得d=2，D=4
所以an=2n+1,bn...

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设An公差为d，Bn公差为D
An=(a1+an)n/2
Bn=(b1+bn)n/2
An/Bn=(a1+an)/(b1+bn)=[6+(n-1)d]/[4+(n-1)D]
因为limAn/Bn=1/2，所以的d/D=1/2,D=2d
因为b2是a2与a3的等差中项，所以a1+d+a1+2d=2(b1+D),得d=2，D=4
所以an=2n+1,bn=4n-2
所以lim[1/(a1b1)+1/(a2b2)+……+1/(anbn)]=lim[2/2*6+2/6*10+2/10*14+...2/(4n-2)(4n+2)]
=lim2*1/4*[1/2-1/6+1/6-1/10+1/10-1/14+...1/(4n-2)-1/(4n+2)]=lim[n/(4n+2)]=1/4

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{An}公差为d，{Bn}公差为d'
An=[2a1+(n-1)d]*n/2
Bn=[2b1+(n-1)d']*n/2
lim An/Bn=lim [6+(n-1)d]/[4+(n-1)d']=lim [6/(n-1)+d]/[4/(n-1)+d']=d/d'
因为limAn/Bn=1/2，所以d/d'=1/2, d'=2d (1)
...

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{An}公差为d，{Bn}公差为d'
An=[2a1+(n-1)d]*n/2
Bn=[2b1+(n-1)d']*n/2
lim An/Bn=lim [6+(n-1)d]/[4+(n-1)d']=lim [6/(n-1)+d]/[4/(n-1)+d']=d/d'
因为limAn/Bn=1/2，所以d/d'=1/2, d'=2d (1)
因为b2是a2与a3的等差中项，所以a1+d+a1+2d=2(b1+d'),
2*3+3d=2*2+2d' 2d'=3d+2 (2)
联立(1)(2) 解得d=2，d'=4
所以an=a1+(n-1)*2=2n+1, bn=b1+(n-1)*4=4n-2=2(2n-1)
所以lim[1/(a1b1)+1/(a2b2)+……+1/(anbn)]
=lim(1/2)[1/1*3+1/3*5+.......+1/(2n-1)(2n+1)]
=lim(1/2)*{(1/2)(1-1/3)+(1/2)(1/3-1/5)+(1/2)[1/(2n-1)-1/(2n+1)]}
=lim(1/2)*(1/2)[1-1/3+1/3-1/5+......+1/(2n-1)-1/(2n+1)]
=lim(1/4)*[2n/(2n+1)]
=1/4

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