已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13 〔1〕求数列...已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13〔1〕求数列{An}和{Bn}的通项公式;〔2〕求{Bn/2An}的前n项和Sn
问题描述:
已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13 〔1〕求数列...
已知数列{An}是首项a1=1的等比数列,且An〉0,{bn}是首项为1的等差数列,又a5+b3=21,a3+b5=13
〔1〕求数列{An}和{Bn}的通项公式;
〔2〕求{Bn/2An}的前n项和Sn
答
(1)设An=q^(n-1),Bn=1+(n-1)d.由所给等式得q^4+2d=20,q^2+4d=12,设m=q^2,n=2d,则m,q>0,简化式子,易解得q=b=2.得An,Bn.(2)Bn/2An=(2n-1)/(2^n).写出Sn的计算式,再写出1/2Sn的计算式,相减得1/2Sn=1/2+1/2^1+1/2^2+…1/...