求(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1的个位数是几?
1个回答
在前面加一个(2-1).
剩下的请自己想,不会再补充.
相关问题
求(1+1/2^2)(1+1/2^4 )(1+2/2^8)(1+2/2^16)(1+2/2^32)+1的个位数字
求值:(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)-2^32
(2+1)(2∧2+1)(2∧4+1)(2∧8+1)(2∧16+1)(2∧32+1)-2∧64求值
(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1=
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)-2^32
2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)的个位数是什么?
[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)的解