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
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