关于数学巧算(1)(3+1)(3^2+1)(3^4+1)(3^5+1)...(3^2n+1)(2) (1+1/2)(1+

问题描述:

关于数学巧算
(1)(3+1)(3^2+1)(3^4+1)(3^5+1)...(3^2n+1)
(2) (1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
1个回答 分类:数学 2014-09-24

问题解答:

我来补答
(3+1)(3^2+1)(3^4+1)(3^5+1)……(3^2n+1)
=(3-1)(3+1)(3^2+1)(3^4+1)(3^5+1)……(3^2n+1)/2
=(3^2-1)(3^2+1)(3^4+1)(3^5+1)……(3^2n+1)/2
=……
=(3^(2n+1)+1)/2
2)
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=(1-1/2)(1+1/2)(1+1/2的2次方)(1+1/2的4次方)(1+1/2的8次方)/(1-1/2)+1/2的15次方
=(1-1/2的2次方)(1+1/2的2次方)(1+1/2的4次方)(1+1/2的8次方)/(1-1/2)+1/2的15次方
=.
=(1-1/2的8次方)(1+1/2的8次方)/(1-1/2)+1/2的15次方
=(1-1/2的16次方)/(1/2)+1/2的15次方
=2-1/2的15次方+1/2的15次方
=2
 
 
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