an是等差数列,bn是各项都为正数的等比数列,a1=b1,a3+b5=21,a5+b3=13,求an乘bn的前n项和sn

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设an=1+(n-1)d,bn=q^(n-1)
则1+2d+q^4=21
1+4d+q^2=13
解得d=2
q=2
所以 an=2n-1,bn=2^(n-1)
令Sn为数列an/bn的前项和,则
Sn=1/1+3/2+5/4+……+(2n-1)/2^(n-1) ①
两边同乘以2,得
2Sn=2/1+3/1+5/2+……+(2n-1)/2^(n-2) ②
②-①【错位相减】,得
Sn=2/1+2/1+2/2+……+2/2^(n-2)-(2n-1)/2^(n-1)
=2+2*［1/1+1/2+1/4+……+1/2^(n-2)］-(2n-1)/2^(n-1)
=2+2*［1-(1/2)^(n-1)］/(1-1/2)- (2n-1)/2^(n-1)
=2+4*［1-(1/2)^(n-1)］-(2n-1)/2^(n-1)
=2+4-4/2^(n-1)-(2n-1)/2^(n-1)
=6-(4+2n-1)/2^(n-1)
=6-(2n+3)/2^(n-1)
再问：不是an除以bn啊，是an乘以bn
再答：啊？看错了，汗……，我重做……
再问：嗯嗯。
再答：令Sn为数列an*bn的前项和，则 Sn=1*1+3*2+5*4+……+(2n-1)*2^(n-1) ① 两边同乘以2，得 2Sn=1*2+3*4+5*8+……+(2n-1)*2^n ② ① - ②【错位相减】，得 -Sn=1*1+2*2+2*4+……+2*2^(n-1)-(2n-1)*2^n =1+2*2*［2^(n-1)-1］/(2-1) -(2n-1)*2^n =1+2^(n+1)-4-(2n-1)*2^n Sn=(2n-1)*2^n -2^(n+1)+3 =(2n-1-2)2^n+3 =(2n-3)2^n+3

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