(1-1/2²)(1-1/3²)…(1-1/2011²)(1-1/2012²)

(1-1/2²)(1-1/3²)…(1-1/2011²)(1-1/2012²)
(1-1/2²)(1-1/3²)…(1-1/2011²)(1-1/2012²)

(1-1/2²)(1-1/3²)…(1-1/2011²)(1-1/2012²)
(1-1/2^2)(1-1/3^2)(1-1/4^2)···(1-1/n^2)···(1-1/100^2)
=(i=2,100)∏(1-1/i^2)
=(i=2,100)∏[(1+1/i)(1-1/i)]
=(i=2,100)∏(1+1/i)×(i=2,100)∏(1-1/i)
=[(3/2)×(4/3)×(5/4)×···×(101/100)]
×[(1/2)×(2/3)×(3/4)×···×(99/100)]
=(101/2)×(1/100)
=101/200
____∏是求积符合,和求和符号∑是一样的!用∏,可以节约很多笔墨,思路清晰,不易出错,用它好啊

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