先利用平方差公式，过程如下：
an=(1/4)(an + 1)^2 - (1/4)(an-1 +1)^2
=(1/4)((an + 1)^2 - (an-1 +1)^2)
=(1/4)(an + 1 + an-1 +1)(an + 1 - (an-1 +1))
=(1/4)(an + an-1 +2)(an - an-1)
=(1/4)[(an + an-1)(an - an-1) + 2(an - an-1)]
等式两边同时乘以4，得到
4an = (an + an-1)(an - an-1) + 2(an - an-1)
移项，得
0 = (an + an-1)(an - an-1) + 2(an - an-1) -4an
0 = (an + an-1)(an - an-1) + 2(-an - an-1)
0 = (an + an-1)(an - an-1) - 2(an + an-1)
0 = (an + an-1)(an - an-1 -2)
这是平方差公式啊
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