\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

= \(1-\frac{1}{50}

Ta có : 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

= 1 - 1/50 < 1

Nên 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 < 1

1/1.2   +  1/2.3   +  1/3.4  + ... +   1/49.50

=1-1/2+1/2-1/3+...+1/49-1/50

=1-1/50

=49/50 

ta có :9/10=45/50

=>49/50>45/50

=>1/1.2   +  1/2.3   +  1/3.4  + ... +   1/49.50  > 9/10

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

= 1 - 1/50

= 49/50

=> 49/50 > 9/10

=> 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 > 9/10

Chúc bn có kết quả hc kì II thật tốt nha !!!!!!!!!!!! ^_^

mình làm luôn nha :

    Ta có : 1/1.2+1/2.3+1/3.4+.....+1/49.50 =1-1/2+1/2-1/3=1/3-1/4+....+1/49-1/50

                                                             =1-1/50=49/50

Vì 49/50>45/50 = 9/10 nên 45/50>9/10

                                     Vậy 1/1.2+1/2.3+1/3.4+...+1/49.50 > 9/10

\(S_1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(S_2=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{94\cdot97}+\frac{1}{97\cdot100}\)

\(3S_2=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+....+\frac{3}{94\cdot97}+\frac{3}{97\cdot100}\)

\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{97}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)

=> \(S_2=\frac{6}{25}:3=\frac{2}{25}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}< 1\) (đpcm)

ta có :

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}< 1\)

thanks mấy bạn nhiều lắm ^^

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

=\(1-\frac{1}{50}\)

Vì \(1-\frac{1}{50}< 1\)nên A < 1

B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=\(\frac{1}{2}-\frac{1}{100}\)

Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(\Rightarrow A< 1\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=\frac{1}{2}-\frac{1}{100}\)

\(\Rightarrow B< \frac{1}{2}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}< 1\)

\(B=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2003\cdot2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ...+ \(\frac{1}{2003.2004}\)

= 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+...+ \(\frac{1}{2003}\)- \(\frac{1}{2004}\)

= 1 - \(\frac{1}{2004}\)

= \(\frac{2003}{2004}\)

= 

M=1-1/2+1/2-1/3+...+1/49-1/50

=1-1/50<1

\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{49.50}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)

\(M=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+........+\left(-\frac{1}{49}+\frac{1}{49}\right)-\frac{1}{50}\)

\(M=\frac{1}{1}-0+0+0+0+0+......+0+0-\frac{1}{50}\)

\(M=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)

Vì \(\frac{49}{50}<1\) nên  \(S<1\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

     \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

     \(=1-\frac{1}{50}<1\)

\(\Rightarrow M<1\) 

Vậy \(M<1\)

Chúc bạn học tốt!!!!!!!

M=1/1.2+1/2.3+1/3.4+...+1/49.50

M=1-1/2+1/2-1/3+...+1/49-1/50

M=1-1/50<1

Vậy M<1