计算题(1+1/2)(1+1/4)(1+1/16)(1+1/256)…(1+1/2的2n次方)

问题描述:

计算题(1+1/2)(1+1/4)(1+1/16)(1+1/256)…(1+1/2的2n次方)
请利用平方差、完全平方公式及其变形公式得来,要有详细过程.谢!
1个回答 分类:数学 2014-11-02

问题解答:

我来补答
(1+1/2)(1+1/4)(1+1/16)(1+1/256)…(1+1/1的2n-2次方)1+1/2的2n次方)
先乘以(1-1/2),整个式子再除以(1-1/2)
=(1-1/2)(1+1/2)(1+1/4)(1+1/16)…(1+1/2的2n+2次方)(1+1/2的2n次方)/(1/2)
=(1-1/4)(1+1/4)(1+1/16)…(1+1/2的2n+2次方)(1+1/2的2n次方)*2
=(1-16)*(1+1/16)*(1+1/256)……(1+1/2的2n+2次方)(1+1/2的2n次方)*2
=(1-1/2的2n次方)(1+1/2的2n次方)*2
=(1-1/2的4n次方)*2
=2-1/2的(4n-1)次方
 
 
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