已知递增数列an满足a1=6,且an+a(n-1)=9/(an-a(n-1))+8(n>=2),则a70=?

已知递增数列an满足a1=6,且an+a(n-1)=9/(an-a(n-1))+8(n>=2),则a70=?
已知递增数列an满足a1=6,且an+a(n-1)=9/(an-a(n-1))+8(n>=2),则a70=?

已知递增数列an满足a1=6,且an+a(n-1)=9/(an-a(n-1))+8(n>=2),则a70=?
变形可得(an^2-8an)-(an-1^2-8an-1)=9
构造数列bn=an^2-8an,则bn为一等差数列,公差为9,首项为b1=a1^2-8a1=-12
则b70=-12+9*69=609
即a70^2-8*a70=609
(a70+21)(a70-29)=0
a70=-21或a70=29
又an为递增数列,a70>a1=6,
故a70=29