已知函数f(x)=x+4除以x+1,数列an中,a1=1,an+1=f(an),n属于正整数

1个回答

  • 1.

    x=f(x)

    x=(x+4)/(x+1)

    x²=4

    x=2或x=-2

    m=2 n=-2或m=-2 n=2

    a(n+1)=f(an)=(an+4)/(an +1)

    a(n+1)+2=(an+4+2an+2)/(an+1)=3(an+2)/(an+1)

    a(n+1)-2=(an+4-2an-2)/(an+1)=-(an-2)/(an+1)

    [a(n+1)+2]/[a(n+1)-2]=(-3)[(an+2)/(an-2)]

    {[a(n+1)+2]/[a(n+1)-2]}/[(an+2)/(an-2)]=-3,为定值.

    {[a(n+1)-2]/[a(n+1)+2]}/[(an-2)/(an+2)]=-1/3,为定值.

    (a1+2)/(a1-2)=(1+2)/(1-2)=-3

    数列{(an+2)/(an-2)}是以-3为首项,-3为公比的等比数列.

    (a1-2)/(a1+2)=(1-2)/(1+2)=-1/3

    数列{(an-2)/(an+2)}是以-1/3为首项,-1/3为公比的等比数列.

    m=-2 n=2时,(an-m)/(an-n)是以-3为首项,-3为公比的等比数列.

    m=2 n=-2时,(an-m)/(an-n)是以-1/3为首项,-1/3为公比的等比数列.

    2.

    (an+2)/(an-2)=(-3)ⁿ

    an=2[(-3)ⁿ+1]/[(-3)ⁿ-1]=2[(-3)ⁿ-1+2]/[(-3)ⁿ-1]=2 +4/[(-3)ⁿ-1]

    数列{an}的通项公式为an=2+ 4/[(-3)ⁿ-1]

    n为奇数时,(-3)ⁿ2