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

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

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

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

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

Vậy M < 1

2A = 1 + 1/2 + 1/2² + 1/2³ + ... + 1/2⁴⁸+ 1/2⁴⁹

2A - A = (1 + 1/2 + 1/2² + 1/2³ + ... + 1/2⁴⁸ + 1/2⁴⁹) - (1/2 + 1/2² + 1/2³ + 1/2⁴ + ... + 1/2⁴⁹ + 1/2⁵⁰)

A = 1 - 1/2⁵⁰

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

\(2A=2\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(2A=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\)

\(2A-A=A\)

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

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

\(=1-\frac{1}{2^{50}}< 1\)

\(\Rightarrow A< 1\)

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

          \(2A=\text{​​}\text{​​}1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

             \(A=1-\frac{1}{2^{50}}\)

             Vậy \(A\)<  1

1-1/2^50 sao lại bé hơn 1 vậy :V?

2A=1+1/2+................+1/2^49+1/2^50

A=1+1/2^50=> A>1

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

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

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{49}{50}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}\)

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

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

= \(\left(1-\dfrac{1}{50}\right)\) = \(\dfrac{49}{50}\)

#)Thắc mắc ? 

Cho mk hỏi cái ''với 2'' là j bn ? so sánh ak, nếu là so sánh thì mk giải thế này :

#)Giải :

\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)

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

\(M=2-\frac{2}{50}\)

\(M=1\frac{24}{25}=\frac{49}{25}\)

So sánh \(\frac{49}{25}\)với  2

\(2=\frac{2}{1}=\frac{50}{25}\)

Vì \(\frac{49}{25}< \frac{50}{25}\Rightarrow\frac{49}{25}< 2\Rightarrow M< 2\)

          #~Will~be~Pens~#

\(M=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{49.50}=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=2\left(1-\frac{1}{50}\right)=2x\frac{49}{50}=\frac{49}{25}=1\frac{24}{25}\)

Vì M=1 24/25

=>M<2

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

          \(B=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

          \(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)

            \(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)\)

             \(B=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)

\(\Rightarrow\)\(B=A\)