(1)(1+1/2)(1+1/2的二次方)(1+1/2的四次方)(1+1/2的八次方)(1+1/2的十六次方)(2)(3+1)(3的二次方+1)(3的四次方+1).(3的2次方的n次方+1)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 20:14:44
![(1)(1+1/2)(1+1/2的二次方)(1+1/2的四次方)(1+1/2的八次方)(1+1/2的十六次方)(2)(3+1)(3的二次方+1)(3的四次方+1).(3的2次方的n次方+1)](/uploads/image/z/435041-17-1.jpg?t=%281%29%281%2B1%2F2%29%281%2B1%2F2%E7%9A%84%E4%BA%8C%E6%AC%A1%E6%96%B9%29%281%2B1%2F2%E7%9A%84%E5%9B%9B%E6%AC%A1%E6%96%B9%29%281%2B1%2F2%E7%9A%84%E5%85%AB%E6%AC%A1%E6%96%B9%29%281%2B1%2F2%E7%9A%84%E5%8D%81%E5%85%AD%E6%AC%A1%E6%96%B9%29%282%29%283%2B1%29%283%E7%9A%84%E4%BA%8C%E6%AC%A1%E6%96%B9%2B1%29%283%E7%9A%84%E5%9B%9B%E6%AC%A1%E6%96%B9%2B1%29.%283%E7%9A%842%E6%AC%A1%E6%96%B9%E7%9A%84n%E6%AC%A1%E6%96%B9%2B1%29)
(1)(1+1/2)(1+1/2的二次方)(1+1/2的四次方)(1+1/2的八次方)(1+1/2的十六次方)(2)(3+1)(3的二次方+1)(3的四次方+1).(3的2次方的n次方+1)
(1)(1+1/2)(1+1/2的二次方)(1+1/2的四次方)(1+1/2的八次方)(1+1/2的十六次方)
(2)(3+1)(3的二次方+1)(3的四次方+1).(3的2次方的n次方+1)
(1)(1+1/2)(1+1/2的二次方)(1+1/2的四次方)(1+1/2的八次方)(1+1/2的十六次方)(2)(3+1)(3的二次方+1)(3的四次方+1).(3的2次方的n次方+1)
(1)(1+1/2)(1+1/2的二次方)(1+1/2的四次方)(1+1/2的八次方)(1+1/2的十六次方)
=2*(1-1/2)*)(1+1/2)(1+1/2的二次方)(1+1/2的四次方)(1+1/2的八次方)(1+1/2的十六次方)
=2(1-1/2^2)(1+1/2的二次方)(1+1/2的四次方)(1+1/2的八次方)(1+1/2的十六次方)
=2*(1-1/2^4)(1+1/2的八次方)(1+1/2的十六次方)
.
=2*(1-1/2^32)
=2-1/2^31
(2)(3+1)(3的二次方+1)(3的四次方+1).(3的2次方的n次方+1)
=1/2*(3-1)(3+1)(3的二次方+1)(3的四次方+1).(3的2次方的n次方+1)
=1/2*(3^2-1)(3的四次方+1).(3的2次方的n次方+1)
...
=1/2*(3^(4n)-1)
=[3^(4n)-1]/2
平方差公式应该学过吧,这两个在一种解法,第一个前面乘以(1-1/2)然后用平方差公式化简。最后再除以(1-1/2)
第二个乘以(3-1)最后除(3-1)
a^2-b^2=(a-b)(a+b)的应用,配上一项(a-b)就可以搞定
2*(1+1/2)(1-1/2)[1+(1/2)^2][1+(1/2)^4][1+(1/2)^8][1+(1/2)^16]=2*[1-(1/2)^32]=1.9999999