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