Question

a1=√2,an^2=2+an-1求an

a1=根号2，an=根号（2+an-1),求an通项

Answer 1

a1=√2=2cos(π/庆岁败誉颤4);
a2=√雀此(2+2cos(π/4))=2cos(π/8);.........,
an=2cos[π/(2^(n+1))].

Answer 2

2(an)^2=[a(n+1)]^2+[a(n-1)]^2
[a(n+1)]^2-[an]^2 = (an)^2 - [a(n-1)]^2
= (a2)^2- (a1)^2
= 3
[a(n+1)]^2-[an]^2 =3
(an)^2-(a2)^2 = 3(n-2)
(an)^2 = 3n-2
an = √(3n-2)
这样可链神以么做清？棚胡亏

追问

n=1时不对，通项完全不对