\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)

\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)

\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)

\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)

\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)

\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)

\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)

\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)

\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)

Giải:

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}.\)

\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{29.31}{30^2}.\)

\(A=\dfrac{1.2.3.....29}{2.3.4.....30}.\dfrac{2.3.4.....31}{2.3.4.....30}.\)

\(A=\dfrac{1}{30}.31=\dfrac{30}{31}.\)

Vậy \(A=\dfrac{30}{31}.\)

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}............................\dfrac{899}{30^2}\)

\(\Leftrightarrow A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}..............................\dfrac{29.31}{30^2}\)

\(\Leftrightarrow A=\dfrac{1.3.2.4.3.5..........29.31}{2.2.3.3.4.4.........30.30}\)

\(\Leftrightarrow A=\dfrac{\left(2.3.........29.30\right).\left(3.4.5......29.31\right)}{\left(2.3....29.30\right).\left(2.3.4.......29.30\right)}\)

\(\Leftrightarrow A=\dfrac{31}{2.30}=\dfrac{31}{60}\)

\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}....................................\dfrac{2499}{2500}\)

\(\Leftrightarrow B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.............................\dfrac{49.51}{50.50}\)

\(\Leftrightarrow B=\dfrac{\left(2.3.4.....49\right).\left(4.5.6......51\right)}{\left(3.4.5....50\right)\left(3.4.5.....50\right)}=\dfrac{2.51}{50.3}=\dfrac{17}{25}\)

a)\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{100^2-1}\)

\(A< \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)

\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}< \dfrac{50}{100}=\dfrac{1}{2}\)

Vậy \(A< \dfrac{1}{2}\)

b)B=\(\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{2499}{2500}\)

49-B=\(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

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

\(49-B< 1-\dfrac{1}{50}< 1\Leftrightarrow49< 1+B\Leftrightarrow B>48\)(ĐPCM)

b) Đặt :

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+............+\dfrac{2499}{2500}\)

\(\Rightarrow A=\dfrac{4}{4}-\dfrac{1}{4}+\dfrac{9}{9}-\dfrac{1}{9}+.........+\dfrac{2500}{2500}-\dfrac{1}{2500}\)

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

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

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

Lại có :

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)

Mà :

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

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

\(=1-\dfrac{1}{50}< 1\)

\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)>-1\)

\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{50^2}\right)>49-1\)\(=48\)

\(\Rightarrow A>48\) \(\rightarrowđpcm\)

sai đề bn ơi

54 là 51 mới đúng

với lại lũy thừa tất cả phải là mũ 2

\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{50^2}{49.51}\\ =\dfrac{\left(2.3.4.....50\right).\left(2.3.4....50\right)}{\left(1.2.3....49\right).\left(3.4.5.....51\right)}\\ =\dfrac{50.2}{51.1}\\ =1\dfrac{49}{51}\\ =\dfrac{100}{51}\)

\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)

Từ đó ta có:

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

\(\Rightarrow A>49-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)

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

\(\Rightarrow A>49-\left(1-\dfrac{1}{50}\right)=48+\dfrac{1}{50}>48\)

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\\ A=\left(1+1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\\ A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\)

Có \(\dfrac{1}{4}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\\ \dfrac{1}{9}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\\ \dfrac{1}{16}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\\ ...\\ \dfrac{1}{2500}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1-\dfrac{1}{50}< 1\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1\)

\(\Rightarrow A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)>49-1\\ \Rightarrow A>48\)

a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)

b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)

A bn lướt xuống dưới mà xem cách làm 

nhưng của bn là cho 3 ra ngoài nha

Giải:

A=3/1.3+3/3.5+3/5.7+...+3/49.51

A=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)

A=3/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51)

A=3/2.(1/1-1/51)

A=3/2.50/51

A=25/17

B=1/3+1/3²+1/3³+...+1/3⁸

3B=1+1/3+1/3²+...+1/3⁷

3B-B=(1+1/3+1/3²+...+1/3⁷)-(1/3+1/3²+1/3³+...+1/3⁸)

2B=1-1/3⁸

   B=1-1/3⁸ /2

Chúc bạn học tốt!