高一数列求通项公式

因为[a（n+1）/an]^2+[a（n+1）/a(n+2)]^2=2
两边同时除以[1/a(n+1)]^2得：
[1/an]^2+[1/a(n+2)]^2=2/a(n+1)^2
即：[1/an]^2+[1/a(n+2)]^2=1/a(n+1)^2+1/a(n+1)^2
移项[1/a(n+2)]^2-1/a(n+1)^2=1/a(n+1)^2-[1/an]^2
数列{1/a(n+1)^2-[1/an]^2}是常数列
即有1/a(n+1)^2-[1/an]^2=(1/a2)^2-(1/a1)^=3/4
即有数列{[1/an]^2}是等差数列，其首项为[1/a1]^2=1/4，公差为3/4
即有[1/an]^2=1/4+3n/4=(3n+1)/4
n>0
两边开方即可

都除以[a(n+1)]^2
得[1/an]^2 +[a(n+2)]^2 = 2[a(n+1)}^2
[1/an]^2就是等差数列
[1/an]^2=0.75n-0.5
an=...