(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^32+1)的个位数字

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(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)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)

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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

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1/2,1/4, (1-1/2^2)(1-1/3^2)(1-1/4^2).(1-1/2009^2), (2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1).({2}^{64}+1)+1 巧算((2^1+1)(2^2+1)(2^4+1)(2^8+1)+1)/2^15 计算(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1 计算:(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16.计算：(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16. [(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)] (1-1/2^2)*(1-1/3^2)*(1-1/4^2)*.*(1-1/2002^2)*(1-1/2003^2) (1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16), 1,2,4,4,1,( ) (2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1 (2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1） (2+1)(2-1)(2^2+1)(2^4+1).(2^8+1)化简 (1+2)*(1+2^2)*(1+2^4)*(1+2^8)*(1+2^16) 化简(1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16) 化简(2+1)(2^2+1)(2^4+1)(2^8+1)…(2^256+1) (2-1)(2+1)(2^2+1)(2^4)...(2^64+1)+1=? (2+1)(2^2+1)(2^4+1)(2^8+1)……(2^1024+1)