(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)的个位数字
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 10:16:35
![(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)的个位数字](/uploads/image/z/2985861-21-1.jpg?t=%282%2B1%29%282%5E2%2B1%29%282%5E4%2B1%29%282%5E8%2B1%29%E2%80%A6%E2%80%A6%282%5E32%2B1%29%E7%9A%84%E4%B8%AA%E4%BD%8D%E6%95%B0%E5%AD%97)
(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)的个位数字
(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)的个位数字
(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)的个位数字
(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)
=(2^4-1)(2^4+1)(2^8+1)……(2^32+1)
=(2^8-1)(2^8+1)……(2^32+1)
=(2^16-1)(2^16+1)(2^32+1)
=(2^32-1)(2^32+1)
=2^64-1
因为2^n的个位数有变化规律:2,4,8,6,2,4,8,6,.周期是4
那么2^64的个位数与2^4相同
为6
那么2^64-1的个位数是5
如果不懂,祝学习愉快!
这是一道渐渐化简的题目,关键是想到第一步即2+1=(2+1)(2-1)=3(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)=(2+1)(2-1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)
=(2^4-1)(2^4+1)(2^8+1)……(2^32+1)
全部展开
这是一道渐渐化简的题目,关键是想到第一步即2+1=(2+1)(2-1)=3(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)=(2+1)(2-1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)
=(2^4-1)(2^4+1)(2^8+1)……(2^32+1)
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)
=(2^16-1)(2^16+1)(2^32+1)
=(2^32-1)(2^32+1)
=2^64-1
收起