数列an定义为a1=a2=1,an+2=an+1+an,求证：当n>=2时,a2n-1必是数列中某两项平方和,a2n必是

【证明】
此数列为著名的斐波拉契数列
数列的前4项为1,1,2,3,
a3=a1^2+a2^2
a4=a3^2-a1^2
设 a2n-1=an-1^2+an^2
a2n=an+1^2-an-1^2 (n>=2)
两式相加即得：
a2n+1=an^2+an+1^2
而：a2n+2=a2n+2an+1
=2an+1^2+an^2-an-1^2
=2an+1^2+2an^2-an^2-an-1^2
=（an+1+an)^2+(an+1-an)^2-an^2-an-1^2
=an+2^2+an-1^2-an^2-an-1^2
=an+2^2-an^2
因此对一切自然数n≥2,
a2n-1=an-1^2+an^2
a2n=an+1^2-an-1^2

看图片

满意答案中的a2n+2=a2n+2an+1
应变为a2n+2=a2n+a2n+1