数学好的伙伴儿们帮个忙吧！
\u6570\u5b66\u597d\u7684\u4f19\u4f34\u513f\u4eec\u5e2e\u4e2a\u5fd9\u5427\uff01\u628ax=0\uff0cY=0\u548cx=1\uff0cY=2\u4ee3\u5165Y=kx+b
\u5f97\u4e00\u4e8c\u5143\u4e00\u6b21\u65b9\u7a0b\u7ec4\uff1a0=b\uff0c2=k+b
\u89e3\u5f97b=0\uff0ck=2
\u628ab=0\uff0ck=2\u4ee3\u5165Y=kx+b
\u5f97Y=2x
\u5f53Y>0\u65f6
2x>0
x>0
[1 - (1/2)²] = (1+1/2)(1-1/2) = 1/2 * 3/2
[1 - (1/3)²] = (1+1/3)(1-1/3) = 2/3 * 4/3
[1 - (1/4)²] = (1+1/4)(1-1/4) = 3/4 * 5/4
[1 - (1/5)²] = (1+1/5)(1-1/5) = 4/5 * 6/5
..........
[1 - (1/2011)²] = (1+1/2011)(1-1/2011) = 2010/2011 * 2012/2011
[1 - (1/2012)²] = (1+1/2012)(1-1/2012) = 2011/2012 * 2013/2012
相乘：
(1/2 * 3/2 )* (2/3 * 5/3)* (3/4 * 5/4)* (4/5 * 6/5) *.......* (2012/2011 * 2012/2011)* (2011/2012 * 2013/2012) =
1/2* (3/2 *2/3) * (4/3* 3/4)* (5/4 * 4/5) *(6/5*5/6)*......*(2012/2011 * 2011/2012) * 2013/2012 =
1/2 * 1 * 1*...........* 1 *2013/2012 =
2013 / 4024
平方差分解
1/2*3/2*2/3*4/3*3/4.....2013/2012
=1/2*2013/2012
=2013/4024
顺道求采纳
原式=（1-1/2）（1+1/2）*（1-1/3）（1+1/3）.....
=1/2 * 3/2 * 2/3 * 4/3........
=1/2 * 2013/2012最后一步不用我说了吧~
好像得
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