一道求数列问题,a1=1,a(n+1)+2a^2 n ,求a n

a
= 2(a)^2
= 2*[2(a)^2]^2 = 2^3 * (a)^4
= 2^3*[2(a)^2]^4 = 2^7 * (a)^8
= 2^7*[2(a)^2]^8 = 2^15 * (a)^16
=2^15*[2(a)^2]^16 = 2^31 * (a)^32
= ……
= 2^[2^(n-1) -1] * (a1)^[2^(n-1)]
= 2^[2^(n-1) -1]
内的数字 是下角标
检验
n= 1
a1 = 1
n=2
a2 = 2^(2^1 -1) = 2 成立
n =3
a3 = 2^(2^2 -1) = 8 成立
n=4
a4 = 2^(2^3 -1) = 2^7 = 128 成立

a1=1,a(n+1)=2a^2 n
假设n=1
则a1=2a^2=1,得到a=√2/2
然后再假设n=n-1,求得
a(n)=√2^( n-1)