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。
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![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。](/uploads/image/z/934991-71-1.jpg?t=%E6%9C%89%E4%B8%A4%E4%B8%AA%E5%90%84%E9%A1%B9%E9%83%BD%E6%98%AF%E6%AD%A3%E6%95%B0%E7%9A%84%E6%95%B0%E5%88%97an%2Cbn%2C%E5%A6%82%E6%9E%9Ca1%3D1%2Cb1%3D2%2Ca2%3D3%E4%B8%94an%2Cbn%2Can%2B1%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97bn%2Can%2B1%2Cbn%2B1%E6%88%90%E7%AD%89%E6%AF%94%EF%BC%8C%E6%B1%82%E8%BF%99%E4%B8%A4%E4%B8%AA%E6%95%B0%E5%88%97%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E3%80%82)
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。
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。
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。
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