(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1=
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1=
1/2+1/4+1/8+1/16+1/32+.
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1计算方法是什么?
计算(1+2)(1+2²)(1+2^4)(1+2^8)(1+2^16)
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
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+
计算(1+2)(1+2^2;)(1+2^4)(1+2^8)(1+2^16)(1+2^32)