设｛an｝是等差数列,｛bn｝是各项都为正数的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13.
（一）求｛an｝、｛bn｝的通项公式；（二）求数列｛an／bn｝的前n项和Sn.
人气：261 ℃ 时间：2019-08-17 22:37:58
解答
设an=a1+(n-1)dbn=b1(n-1)^qa1=b1=1.(1)a5+b3=13.(2)a3+b5=21.(3)4d+q^2=12.(4)2d+q^4=20.(5)(5)*2-(4)得2q^4-q^2-28=0(q^2-4)(2q^2+7)=0q^2=4q=2d=2an=2n-1bn=2^(n-1) Sn=1/1+3/2+5/4+...+(2n-1)/2^(n-1)(1/2)S=1/2...
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