Question

在线性代数里面冒出来的求数列通项公式,

Answer 1

a(n+1)=(1+an)/(3-an)

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

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

1/[a(n+1) -1] - 2/3=4/(an -1) -8/3=4[1/(an -1) -2/3]

题目没有给出a1,不好继续写了.