第一题,解
3/2-1/4-1/8-1/16-1/32-1/64-1/126
= 3/2-(1/4+1/81/16+1/32+1/64+1/126)
=3/2- [(32+16+8+4+2+1)/128]
=3/2-63/128
=192/128-63/128
=129/128
=1又1/128
第二题,解
1/2005+2/2005+3/2005+.+2004/2005
=[【（1+2004）×2004】÷2] / 2005
=[（2005×2004）÷2] / 2005
=2005×2004×2004× 1/2 × 1/2005
=1002
第三题,解
401040104010/200520052005
=401040104010÷200520052005
=2