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
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【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

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