已知两个等差数列{an}和{bn}的前n项和分别为An和Bn,

2个回答

  • ∵An/Bn=(7n+45)/(n+3)

    An/Bn=[(a1+an)n/2]/[(b1+bn)n/2]

    =(a1+an)/(b1+bn)

    (a1+an)/(b1+bn)=(7n+45)/(n+3)

    (下面需将an/bn------> [a1+ax]/[b1+bx]形式)

    ∴an/bn=(2an)/(2bn)

    =[a1+a(2n-1)]/[b1+b(2n-1)]

    =[7(2n-1)+45]/[(2n-1)+3]

    =(14n+38)/(2n+2)

    =(7n+19)/(n+1)

    =[7(n+7)+12]/(n+1)

    =7+12/(n+1)

    ∵an/b2n为整数

    ∴n+1=2,3,4,6,12

    ∴n=1,2,3,5,11共5个

    没看清楚,是an/b2n

    ∵An/Bn=(7n+45)/(n+3)

    ∴An=kn(7n+45),Bn=kn(n+3)

    k为常数

    an=An-A(n-1)

    =7kn^2+45kn-7k(n-1)^2 -45k(n-1)

    =14kn+38k

    bn=Bn-B(n-1)

    =kn^2+3kn-k(n-1)^2-3k(n-1)

    =2kn+2k

    an/b2n=(14kn+38k)/(4kn+2k)

    =(7n+19)/(2n+1)

    =[7(2n+1)+31]/[2(2n+1)]

    =7/2+31/(4n+2)

    ∴31/(4n+2)的小数部分为0.5,

    分母应为62,4n+2=62,n=15

    一个

    (4n+2=6,10,14,18,20,24,.,58均不合题意)