(1-1⼀2的2次方)(1-1⼀3的2次方)(1-1⼀4的2次方)..(1-1⼀2003的2次方)

2025-05-17 17:26:12
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回答1:

计算(1-1/2²)(1-1/3²)(1-1/4²).......(1-1/2008²)

 

解:原式=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4).......(1-1/2008)(1+1/2008)

 

=(1/2)(3/2)(2/3)(4/3)(3/4)(5/4)(4/5)(6/5)........(2002/2003)(2002/2003)

 

=(1/2)(2009/2008)

 

=2004/40/8≈0.5

 

 

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