1/2+1/4+1/8+1/31+1/62+1/124+1/248+1/496=

1/2+1/4+1/8+1/16+1/31+1/62+1/124+1/248+1/496 1/2+1/4+1/8+1/16+1/31+1/62+1/124+1/248+1/496 1/2+1/4+1/6+1/8+.1/n |1/4-1/2|+|1/6-1/4|+|1/8-1/6|+```+|1/2008-1/2006| 1/2+1/4+1/8+1/31+1/62+1/124+1/248+1/496= 计算：1/2+1/4+1/8+1/31+1/62+1/124+1/248+1/496 要有过程. 一道平方差数学题(1/2+1)(1/2^2+1)(1/2^4+1)(1/2^8+1)(1/2^16+1)+1/2^31 把4/31+6/31+8/31+10/31+12/31+14/31+16/31改写成乘法算式是是：2*（2+3+4+5+6+7+8）*1/31？ ((1/2)-1)*((1/3)-1)*((1/4)-1)*((1/5)-1)*((1/6)-1)*((1/7)-1)*((1/8)-1)*((1/9)-1)*((1/10)-1) 1+1+1+1+1+1+1+1+1+2+5+4+8+3+6+2+1+4等于多少? 2/1+4/1+4/1+6/1+8/1+16/1+32/1=? 9*(1-1/2)*(1-1/3)*(1-1/4)*(1-1/5)*(1-1/6)*(1-1/7)*(1-1/8)*(1-1/9)怎样简便计算 1/2*4+1/4*6+1/6*8+1/8*10+---+1/198*200 2分之1+4分之1+8分之1+16分之1+31分之1+62分之1+ 4,2,1,1/2,1/4,（ ) ( ) ( )1/8,1/32,1/4 ,1/16( )( ) ( )1/2,1/6,1/12,1/20,1/30,1/42 ( ) ( ) ( ) 2/1 3/1 4/1 5/1 6/1 8/1 10/1 12/1 15/1 20/1 1/4-1/2+1/6-1/4+1/8-1/6+...+1/2008-1/2006计算题计算题1/4-1/2+1/6-1/4+1/8-1/6+...+1/2008-1/2006 求证：1+1/2+1/3+1/4+1/8+1/6+1/7+^+1/(2n-1)>n/2