数列{an}中,a1+a2+a3+a4+a5=8 a6+a7+a8+a9+a10=56 a11+a12+a13+a14+

1个回答

  • 可能是等差数列、等比数列、其它数列,应说明.

    s5=8

    s10=8+56=64

    s15=?

    1.等差数列:

    设an=a+(n-1)d

    sn=na+(1/2)n(n+1)d

    s5=5a+15d=8

    s10=10a+55d=8+56=64

    a=-104/25

    d=48/25

    s15=15a+120d=168

    a11+a12+a13+a14+a15=s15-s10=168-64=104;

    2.等比数列:

    设an=aq^(n-1)

    sn=a(1-q^n)/(1-q)

    s5=a(1-q^5)/(1-q)=8

    s10=a(1-q^10)/(1-q)=56+8=64

    两式相除:

    1+q^5=64/8=8

    q^5=7

    1-q=a(1-q^5)/8

    s15=a(1-q^15)/(1-q)

    =a(1-q^15)/[a(1-q^5)/8]

    =8(1-q^15)/(1-q^5)

    =8(1-7^3)/(1-7)

    =456

    a11+a12+a13+a14+a15

    =s15-s10

    =456-64

    =392

    3.其它数列 :

    无法算.