计算.1+1/2+1/（2²）+1/（2³）+…+1/（2的2004次方）

令s=1+1/2+(1/2)^2+……+(1/2)^2004(等式两边同时乘以1/2 )所以1/2*s=1/2+(1/2)^2+(1/2)^3+……+(1/2)^2004+(1/2)^2005相减,中间相同的抵消s-1/2*s=1/2*s=1-(1/2)^2005所以s=2-2*(1/2)^2005=2-(1/2)^2004不懂请找我...写清楚一点，最好直接等于……等于……这么写[1+1/2+(1/2)^2+……+(1/2)^2004]-(1/2)(1+1/2+(1/2)^2+……+(1/2)^2004)=(1/2)(1+1/2+(1/2)^2+……+(1/2)^2004)=[1+1/2+(1/2)^2+……+(1/2)^2004]-[1/2+(1/2)^2+(1/2)^3+……+(1/2)^2004+(1/2)^2005]=1-(1/2)^2005故1+1/2+(1/2)^2+……+(1/2)^2004=2-2*(1/2)^2005=2-(1/2)^2004