(a(n+1)-an)^2=2(a(n+1)+an) ①
(an-a(n-1))^2=2(an+a(n-1)) ②
①-②,得(a(n+1)-a(n-1))*（a(n+1)-2an+a(n-1)）=2(a(n+1)-a(n-1)) ③
當a(n+1)=a(n-1)時,數(shù)列是個擺動常數(shù)列 an= a1(2k)
a2(2k-1)( k是正整數(shù))
當a(n+1)≠a(n-1)時,③式可變成a(n+1)-2an+a(n-1)=2④
④式可以變成a(n+1)-an=an-a(n-1)+2 設(shè) bn=a(n+1)-an
則bn=b(n-1)+2 bn是等差數(shù)列bn=b1+2(n-1)=a2-a1+2(n-1)=a(n+1)-an
a(n+1)=an+2n+(a2-a1-2) 設(shè)a2-a1-2=p
a(n+1)=an+2n+p ⑤
a(n+1)-an=2n+p⑥
由⑥可以得 an=(an-a(n-1))+.(a2-a1)+a1=(2n+p+2+p)*n/2
把p的值代入即可得 an的通項