(2+1)(2^2+1)(2^4+1)…(2^32+1)的个位是多少
来源:百度知道 编辑:UC知道 时间:2024/06/06 14:16:25
(2+1)(2^2+1)(2^4+1)…(2^32+1)+1的个位是多少
我都算出原来的个位是7了,你自己再+1,不就行了?
个位是8
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(2+1)(2^2+1)(2^4+1)…(2^32+1)
=1*(2+1)(2^2+1)(2^4+1)…(2^32+1)
=(2-1)(2+1)(2^2+1)(2^4+1)…(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)…(2^32+1)
=(2^4-1)(2^4+1)..(2^32+1)
=...
=(2^32-1)(2^32+1)
=2^64-1
2×2=4,4×2=8,8×2=16,6×2=2,每5个2相乘,尾数就是2
2^64 = (2^5)^12 * 2^4
末位数字=2^12*2^4 = 2^16 = 2^5 * 2^5 * 2^5 *2
末位数字=2*2*2*2=8
最后的个位=8-1=7
(1-√2)^2+(√2-1)^2(√2-1)^2+(-√2-1)^2
1( )2( )
1^2+2^2+...+n^2=?
(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)(2^10+1)......(2^2004+1)
2+2^1+2^2+2^3+...+2^2006=?
(2+1)(2*2+1)(2*2*2*2+1)......(2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2+1)的解法
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^32+1)
(2+1)(2^+1)(2^+1)(2^+1)(2^+1)(2^+1)(2^+1)的值为?
1+1大于2
计算(2+1)(2^2+1)(2^4+1)+。。。。+(2^2n+1)