hello

\(M=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{100\cdot101\cdot102}\\ M=\frac{1}{2}\cdot\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{100\cdot101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{100\cdot101}-\frac{1}{101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10302}\right)\\ M=\frac{1}{2}\cdot\left(\frac{5151}{10302}-\frac{1}{10302}\right)\\ M=\frac{1}{2}\cdot\frac{25}{51}\\ M=\frac{25}{102}\\ \Rightarrow M< 1\)

Vậy M < 1

M<1 ok?

\(M=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{100.101.102}\right)\)

\(M=\frac{1}{2}.\left(1-\frac{1}{102}\right)\)

\(M=\frac{101}{204}< 1\left(đpcm\right)\)

Ta có: M=11.2.3  +12.3.4  +13.4.5  +...+1100.101.102  

         M=2.(11.2.3  +12.3.4  +13.4.5  +...+1100.101.102  ).12 

          M=(21.2.3  +22.3.4  +23.4.5  +...+2100.101.102  ).12 

          M=(11.2  -12.3  +12.3  -13.4  +13.4  -14.5 +...+1100.101 −1101.102  ).12 

          M=( 11.2 −1101.102 ).12 

          Mà 11.2 −1101.102 <1

         Và 12 <1 

        =>  (11.2 −1101.102  ) .12  <1

        => M <1

M=1/1x2x3 =1/2x3x4 +1/3x4x5 +..........+1/100x101x102

M=3-1/1x2x3 +4-2/2x3x4+5-3/3x4x5 + ......... +102-100/100x101x102

M=3/1x2x3 -1/1x2x3 +4/2x3x4 -2/2x3x4 +........... + 102/100x101x102 -100/100x101x102

M=1/1x2 -1/2x3 +1/2x3 -1/3x4 +......... + 1/100x101 -1/101x102

M=1/1x2 -1/101x102

M=2575/5151 < 1   suy ra M<1 

Vậy M<1

cách làm như sau

\(C=\frac{2}{2}.\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{98.99}-\frac{1}{99.100}\right]\)

\(C=1\left[\frac{1}{2}-\frac{1}{9900}\right]\)

\(C=\frac{4949}{9900}\)

cần làm ra ko

CÓ CHỨ

1/2-1/2010.2011

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