(2011•上饶模拟)设an是等差数列,bn是各项都为正数的等比数列,且a1=b1=1,a2+b3=a3+b2=7.

1个回答

  • (1)设an的公差为d,bn的公比为q,则依题意有q>0且

    1+d+q2=7

    1+2d+q=7

    解得d=2,q=2.(2分)

    所以an=1+(n-1)d=2n-1,bn=qn-1=2n-1.(4分)

    (2)因为cn=an-2010=2n-2011≥0⇔n≥1005.5,

    所以,当1≤n≤1005时,cn<0,当n≥1006时,cn>0.(6分)

    所以当n=1005时,An取得最小值.(7分)

    (3)

    an

    bn=

    2n−1

    2n−1.Sn=1+

    3

    21+

    5

    22++

    2n−3

    2n−2+

    2n−1

    2n−1①(9分)2Sn=2+3+

    5

    2++

    2n−3

    2n−3+

    2n−1

    2n−2②

    ②-①得Sn=2+2+

    2

    2+

    2

    22++

    2

    2n−2−

    2n−1

    2n−1=2+2×

    1−

    1

    2n−1

    1−

    1

    2−

    2n−1

    2n−1=6−

    2n+3

    2n−1.(12分)