(2+1) (22+1) (24+1) (28+1) (216+1) (232+1)+1
利用平方差公式计算:(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)+1.
计算:(22+1)(24+1)(28+1)……(232+1)
求(2-1)(2+1)(22+1)(24+1)(28+1)…(232+1)+1的个位数字.
化简:(2-1)(2+1)(22+1)(24+1)…(216+1)+1=
(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
1/22+1/23+1/24+1/25+1/26+1/27+1/28+1/29+1/30=?
3(22+1)(24+1)…(232+1)+1计算结果的个位数字是______.
利用平方差计算(2+1)(22+1)(24+1)(28+1)+1=___.
计算:(2+1)(22+1)(24+1)(28+1)…(2256+1)
(2+1)(22+1)(24+1)(28+1)•••••
分数找规律 8/3 8/1 24/1 ( ) 216/1 3/4 2/1 3/1 ( ),27/4,( )