Question

关于数列的问题

Answer 1

1.
a(n+1)=(3an+2)/(an+4)
a(n+1)+k=[(k+3)an+(4k+2)]
a(n+1)-1=(3an+2-an-4)/(an+4)=(2an-2)/(an+4)+2(an-1)/(an+4) (1)
a(n+1)+2=(3an+2+2an+8)/(an+4)=(5an+10)/(an+4)=5(an+2)/(an+4) (2)
(1)/(2)
[a(n+1)-1]/[a(n+1)+2]=(2/5)[(an-1)/(an+2)]
[a(n+1)-1]/[a(n+1)+2]/[(an-1)/(an+2)]=2/5，为定值
(a1-1)/(a1+2)=(4-1)/(4+2)=1/2
数列{(an-1)/(an+2)}是以1/2为首项，2/5为公比的等比数列
(an-1)/(an+2)=(1/2)(2/5)^(n-1)
解得an=(4×5ⁿ+10×2ⁿ)/(4×5ⁿ-5×2ⁿ)
数列{an}的通项公式为an=(4×5ⁿ+10×2ⁿ)/(4×5ⁿ-5×2ⁿ)
2.
a(n+1)=1/(2-an)
a(n+1)-1=(1-2+an)/(2-an)=(an-1)/(2-an)
1/[a(n+1)-1]=(2-an)/(an-1)
=(1-an+1)/(an-1)
=-1+ 1/(an-1)
1/[a(n+1)-1]-1/(an-1)=-1，为定值
1/(a1-1)=1/(4-1)=1/3，数列{1/(an-1)}是以1/3为首项，-1为公差的等差数列
1/(an-1)=1/3 +(-1)(n-1)=(4-3n)/3
an=3/(4-3n) +1=(3n-7)/(3n-4)
数列{an}的通项公式为an=(3n-7)/(3n-4)