题目

有两个各项都是正数的数列an,bn,如果a1=1,b1=2,a2=3且an,bn,an+1成等差数列
bn,an+1,bn+1成等比，求这两个数列通项公式。

提问时间：2020-08-03

答案

由已知得
bn=[an +a(n+1)]/2
a(n+1)²=bn×b(n+1)
=[an+a(n+1)][a(n+1)+a(n+2)]/4
[an +a(n+1)][a(n+1)+a(n+2)]=4a(n+1)²
等式两边同除以a(n+1)²
[an/a(n+1) +1][a(n+2)/a(n+1) +1]=4,为定值.
a2/a1 =3/1
a3/a2 =4/(1+1/3)) -1=2=4/2
假设a(k+1)/ak=(k+2)/k,即ak/a(k+1)=k/(k+2),则
[k/(k+2) +1][a(k+2)/a(k+1) +1]=4
a(k+2)/a(k+1)=4/[k/(k+2) +1] -1
=4/[(2k+2)/(k+2)] -1
=4(k+2)/[2(k+1)] -1
=2(k+2)/(k+1) -1
=(k+3)/(k+1)
=[(k+1)+2]/(k+1),同样满足表达式.
综上,得a(n+1)/an=(n+2)/n
an/a(n-1)=(n+1)/(n-1)
a(n-1)/a(n-2)=n/(n-2)
…………
a2/a1=3/1
连乘
an/a1=(3/1)(4/2)...[(n+1)/(n-1)]=[3×4×...×(n+1)]/[1×2×...×(n-1)]=n(n+1)/2
an=a1n(n+1)/2=n(n+1)/2
bn=[an+a(n+1)]/2=[n(n+1)/2 +(n+1)(n+2)/2]/2=(n+1)²/2
数列{an}的通项公式为an=n(n+1)/2；数列{bn}的通项公式为bn=(n+1)²/2