{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个回答

  • (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]