(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
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(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
![(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?](/uploads/image/z/6078039-15-9.jpg?t=%282%2B1%29%282%5E2%2B1%29%282%5E4%2B1%29%282%5E8%2B1%29%282%5E16%2B1%29%282%5E32%2B1%29%282%5E64%2B1%29%3D%3F)
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)/(2-1)
=(2^2-1)(2^2+1).(2^64+1)
=...
=(2^64-1)(2^64+1)
=(2^128-1)
2^1=2
2^2=4
2^3=8
2^4个位数是6
2^5个位数是2
所以个位数是4个一循环
所以2^128的个位数=2^4的个位数=6
即:2^128-1的个位上是:6-1=5
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)/(2-1)
=(2^2-1)(2^2+1).(2^64+1)
=...
=(2^64-1)(2^64+1)
=(2^128-1)
2^1=2
2^2=4
2^3=8
2^4个位数是6
2^5个位数是2
所以个位数是4个一循环
所以2^128的个位数=2^4的个位数=6
即:2^128-1的个位上是:6-1=5
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
已知m=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128
设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),试确定M-2
设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+
若A=(2+1)(2^2+1)(2^4+1)( 2^8+1)( 2^16+1)( 2^32+1)( 2^64+1) 求A
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
128+64+32+16+8+4+2+1+2/1+4/1+8/1+16/1+32/1=?
求和:1/2+1/4+1/8+1/16+1/32+1/64+.+1/1024=?
1-2/1-4/1-8/1-16/1-32/1-64/1=多少?
1-2/1-4/1-8/1-16/1-32/1-64/1=?简便计算
1/2+1/4+1/8+1/16+1/32+1/64+.+1024=?
1/2+1/4+1/8+1/16+1/32+1/64=?