问几道简算1.1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/2562.1分之1999X2000X(1X2+2X3+3X4+...+1999X2000)第二题应为:1999X2000分之1(1X2+2X3+3X4+...+1999X2000)
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![问几道简算1.1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/2562.1分之1999X2000X(1X2+2X3+3X4+...+1999X2000)第二题应为:1999X2000分之1(1X2+2X3+3X4+...+1999X2000)](/uploads/image/z/1959466-58-6.jpg?t=%E9%97%AE%E5%87%A0%E9%81%93%E7%AE%80%E7%AE%971.1%2F2%2B1%2F4%2B1%2F8%2B1%2F16%2B1%2F32%2B1%2F64%2B1%2F128%2B1%2F2562.1%E5%88%86%E4%B9%8B1999X2000X%281X2%2B2X3%2B3X4%2B...%2B1999X2000%29%E7%AC%AC%E4%BA%8C%E9%A2%98%E5%BA%94%E4%B8%BA%3A1999X2000%E5%88%86%E4%B9%8B1%281X2%2B2X3%2B3X4%2B...%2B1999X2000%29)
问几道简算1.1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/2562.1分之1999X2000X(1X2+2X3+3X4+...+1999X2000)第二题应为:1999X2000分之1(1X2+2X3+3X4+...+1999X2000)
问几道简算
1.1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
2.1分之1999X2000X(1X2+2X3+3X4+...+1999X2000)
第二题应为:1999X2000分之1(1X2+2X3+3X4+...+1999X2000)
问几道简算1.1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/2562.1分之1999X2000X(1X2+2X3+3X4+...+1999X2000)第二题应为:1999X2000分之1(1X2+2X3+3X4+...+1999X2000)
1.等比数列,首项1/2,公比1/2,S8=1/2[1-(1/2)^8]/(1-1/2)=1-(1/2)^8=255/256
1: 255/256
2: (999*2000*255/256)/1
1 1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/256-1/256
=255/256
2看不懂
1) S = 1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
(1/2)*S=1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
上式减下式得
(1/2)*S = 1/2-1/512
S = 255/256
2)请写清楚
S=1X2+2X3+3X4+...
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1) S = 1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
(1/2)*S=1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
上式减下式得
(1/2)*S = 1/2-1/512
S = 255/256
2)请写清楚
S=1X2+2X3+3X4+...+1999X2000
=(1^2+1)+(2^2+2)+...+(1999^2+1999)
=1^2+2^2+3^2+...+1999^2
+1+2+3+...+1999
=
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