(1+1/2)(1+1/22)(1+1/24)(1+1/28)1/2 15
(1+1/2)(1+1/22)(1+1/24)(1+1/28)1/2 15
数学题(1+1/2)(1+1/22)(1+1/24)(1+1/28)+1/215
计算:(2+1)(22+1)(24+1)(28+1)…(2256+1)
利用平方差计算(2+1)(22+1)(24+1)(28+1)+1=___.
求(2-1)(2+1)(22+1)(24+1)(28+1)…(232+1)+1的个位数字.
利用平方差公式计算:(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)+1.
(2+1)(22+1)(24+1)(28+1)•••••
如果x=(1+2)(1+22)(1+24)(1+28)…(1+2256),则x+1是( )
计算:(22+1)(24+1)(28+1)……(232+1)
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
(1)1+1/2+1/3+1/4+1/5+1/6+1/7+1/14+1/28
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?