已知数列an满足（an+1－an）（an+1+an）＝9）,且a1＝2,a＞0.

题目

已知数列an满足（an+1－an）（an+1+an）＝9）,且a1＝2,a＞0.
求证:｛an²｝为等差数列
求｛an｝的通项公式.

答案

a1=2
(a(n+1)－an).(a(n+1)+an)＝9
[a(n+1)]^2- (an)^2 =9
{(an)^2}是等差数列,d=9
(an)^2 - (a1)^2 =9(n-1)
(an)^2 = 9n-5
an = √(9n-5)