Ta có: M=\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +...+\(\frac{1}{100.101.102}\) 

         M=2.(\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +...+\(\frac{1}{100.101.102}\) ).\(\frac{1}{2}\)

          M=(\(\frac{2}{1.2.3}\) +\(\frac{2}{2.3.4}\) +\(\frac{2}{3.4.5}\) +...+\(\frac{2}{100.101.102}\) ).\(\frac{1}{2}\)

          M=(\(\frac{1}{1.2}\) -\(\frac{1}{2.3}\) +\(\frac{1}{2.3}\) -\(\frac{1}{3.4}\) +\(\frac{1}{3.4}\) -\(\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\) ).\(\frac{1}{2}\)

          M=( \(\frac{1}{1.2}-\frac{1}{101.102}\)).\(\frac{1}{2}\)

          Mà \(\frac{1}{1.2}-\frac{1}{101.102}<1\)

         Và \(\frac{1}{2}<1\) 

        \(=>\)  (\(\frac{1}{1.2}-\frac{1}{101.102}\) ) .\(\frac{1}{2}\) \(<1\)

        \(=>\) M <1

Công thức là:1/4.(n-2)(n-1)n(n+1)

=>1.2.3+...+100.101.102=1/4.100.101.102.103

=25.101.102.103

=26527650

Tính ra M to lắm bạn ơi so sánh với 1 đời nào

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

\(\Rightarrow2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{100.101.102}\)

\(\Rightarrow2M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{100.101}-\frac{1}{101.102}\)

\(\Rightarrow2M=\frac{1}{1.2}-\frac{1}{101.102}\)

\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{101.102}\right)=1-\frac{1}{202.102}< 1\)

Vậy M < 1

Anh Kiệt ơi \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)=\frac{1}{4}-\frac{1}{202.102}\)chứ ạ ???

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

Vậy M < 1

M<1 ok?

hello

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

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

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

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

A  = 1.2.3 + 2.3.4 + ....+ 48.49.50

=> 4A = 1.2.3.4 + 2.3.4.(5-1) + ...+ 48.49.50.(51-17)

= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .....+ 48.49.50.51 - 47.48.49.50

= 48.49.50.51

=> A =  48.49.50.51:4 = 12.49.50.51

bài b) làm tương tự nha