已知数列an=[1/a(n-1)]+2,a1=2,求数列通项公式

题目

已知数列an=[1/a(n-1)]+2,a1=2,求数列通项公式

答案

n≥2时,
an=1/a(n-1) +2=[2a(n-1)+1]/a(n-1)
an +√2-1=[(√2+1)a(n-1)+1]/a(n-1)=(√2+1)[a(n-1)+(√2-1)]/a(n-1)
an-√2-1=[(1-√2)a(n-1)+1]/a(n-1)=-(√2-1)[a(n-1)-√2-1]/a(n-1)
(an+√2-1)/(an-√2-1)=-(3+2√2)[a(n-1)+(√2-1)]/[a(n-1)-√2-1]
[(an+√2-1)/(an-√2-1)]/{[[a(n-1)+(√2-1)]/[[a(n-1)-√2-1]}=-(3+2√2),为定值.
(a1+√2-1)/(a1-√2-1)=(2+√2-1)/(2-√2-1)=(√2+1)/(1-√2)=-(3+2√2)
数列{(an+√2-1)/(an-√2-1)}是以-(3+2√2)为首项,-(3+2√2)为公比的等比数列.
(an+√2-1)/(an-√2-1)=[-(3+2√2)]ⁿ
[-(3+2√2)]ⁿan -(√2+1)[-(3+2√2)]ⁿ=an+√2-1
{1-[-(3+2√2)]ⁿ}an=1-√2-(-1)ⁿ·(√2+1)^(2n+1)
an=[1-√2-(-1)ⁿ·(√2+1)^(2n+1)]/{1-[-(3+2√2)]ⁿ}