{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

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{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
(1)求{an},{bn}的通项公式(答案:an=2n-1)
(2)求数列{an/bn}的前n项和Sn(答案:Sn=6-2n+3/2的n次方减一)

/>(1)设{an}的公差为d,{bn}的公比为q
则a3=a1+2d,a5=a1+4d
b3=b1*q^2,b5=b1*q^4
∵a1=b1=1,a3+b5=21,a5+b3=13
∴(1+4d)+q^2=13
(1+2d)+q^4=21
解得d=2,q=2
∴an=1+(n-1)d=2n-1
bn=2^(n-1)[这个是2的n-1次方]
(2)
an/bn=(2n-1)/2^(n-1)
Sn=(2-1)/2^0+(4-1)/2+(6-1)/4+...+(2n-1)/2^(n-1)
=1+3/2+5/4+...+(2n-1)/2^(n-1) ……（1）
2Sn=2+3+5/2+...+(2n-1)/2^(n-2) ……（2）
（2）-（1）得
Sn=2+2+2/2+2/4+...+2/2^(n-2)-(2n-1)/2^(n-1)
=4+1+1/2+...+1/2^(n-3)-(2n-1)/2^(n-1)
=6-(2n+3)/2^(n-1)
注：答案:Sn=6-2n+3/2的n次方减一,有误
正确答案：6-（2n+3)/2^(n-1)[是二的n减一次方]

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