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+b3=9,a5+b2=11.

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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+b3=9,a5+b2=11.
(1) 求{An} {Bn}的通项公式; (2) 求数列{An/Bn}的前n项和Sn.

(1)an=a1+(n-1)d　　　bn=b1*q^(n-1)
带入条件　　　a3+b3=9即为1+2d+q^2=9
a5+b2=11即为1+4d+q=11
联立得2q^2-q-6=0
解得q=-1.5（舍）　q=2 故d=2
an=1+(n-1)d=2n-1 bn=b1*q^(n-1)=2^(n-1)
(2)利用错位相减
用Sn-1/qSn=Sn-0.5Sn=0.5Sn=1+2/2+2/4+……+2/2^(n-1)-(2n-1)/2^n=1+2(1-1/2^(n-1))-(2n-1)/2^n 化简得Sn=6-1/2^(n-3)-(2n-1)/2^(n-1)

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