第1个回答 2012-11-03
1又1/2+2又1/4+3又1/8+...+10又1/1024
=(1+2+3+……10)+(1/2+1/4+1/8+……+1/1024)
=(1+2+3+……10)+[1/2+(1/2)^2+(1/2)^3+……+(1/2)^10]
=10*(10+1)/2+1/2*[1-(1/2)^10]/(1-1/2)
=55+1/2*[1-(1/2)^10]/(1/2)
=55+[1-(1/2)^10]
=56-(1/2)^10
=55+1023/1024
=55又1023/1024
=(1+2+3+……10)+(1/2+1/4+1/8+……+1/1024)
=(1+2+3+……10)+[1/2+(1/2)^2+(1/2)^3+……+(1/2)^10]
=10*(10+1)/2+1/2*[1-(1/2)^10]/(1-1/2)
=55+1/2*[1-(1/2)^10]/(1/2)
=55+[1-(1/2)^10]
=56-(1/2)^10
=55+1023/1024
=55又1023/1024
第2个回答 2012-11-03
等比数列用公式求，公比为3/2
第3个回答 2012-11-03
1又1/2+2又1/4+3又1/8+...+10又1/1024
=（1+2+3+。。。+10）+（1/2+1/4+....+1/1024）
=55+（1/2+1/4+....+1/1024+1/1024）-1/1024
=55+(1/2+1/4+.....+1/512+1/512)-1/1024
.........
=55+(1/2+1/4+1/4)-1/1024
=55+(1/2+1/2)-1/1024
=56-1/1024
=55又1023/1024
=（1+2+3+。。。+10）+（1/2+1/4+....+1/1024）
=55+（1/2+1/4+....+1/1024+1/1024）-1/1024
=55+(1/2+1/4+.....+1/512+1/512)-1/1024
.........
=55+(1/2+1/4+1/4)-1/1024
=55+(1/2+1/2)-1/1024
=56-1/1024
=55又1023/1024
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