baj ney hoj kho do cak p

  • Thread starter justinbieberbelieber0103
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long710200

1,A=3^2/8.11+3^2/11.14+3^2/14.17+...+3^2/197.200
2,chung mjnh:1<1/5+1/6+...+1/16+1/17<2
3,chung mjnh:1/201+1/202+...+1/399+1/400>1/2

1, Ta có:
A=[TEX]\frac{3^{2}}{8.11} + \frac{3^{2}}{11.14} + ... + \frac{3^{2}}{197.200}[/TEX]
= [TEX]3.(\frac{3}{8.11}+\frac{3}{11.14}+ ... + \frac{3}{197.200})[/TEX]
= [TEX]3.(\frac{1}{8}-\frac{1}{11}+ \frac{1}{11}-\frac{1}{14}+ ... + \frac{1}{197}-\frac{1}{200})[/TEX]
= [TEX]3.(\frac{1}{8}-\frac{1}{200})[/TEX]
= [TEX]3.\frac{6}{50}[/TEX]
= [TEX]\frac{9}{25}[/TEX]
3, Các phân số 1/201; 1/202;....;1/399 đều lớn hơn 1/400 nên 1/201+1/202+...+1/399+1/400>1/400 . 200 = 1/2
 
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