\(\Rightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{49.50}\)

\(\Rightarrow A<1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\right)\)

\(\Rightarrow A<1+\left(1-\frac{1}{50}\right)\)

\(\Rightarrow A<1+\frac{49}{50}\)

\(\Rightarrow A<\frac{99}{50}\)

Vì \(\frac{99}{50}<2=\frac{100}{50}\Rightarrow A<2\)  ĐPCM

Ta có:

\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{50^2}<\frac{1}{49.50}\)

Do đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}<1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

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

=>A<2(đpcm)

Em ơi,chứng minh A gì nữa em????

\(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)

\(A=\dfrac{1}{2\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{50\cdot50}\)

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

\(A=1\)

Vậy A=1

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

\(A=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)=1+B\)( Gọi biểu thức trong ngoặc là B)

Ta xét B

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

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

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

B<\(1-\frac{1}{50}<1\)

Vậy B<1

=>A=1+B < 1+1=2

Vậy A<2

đặt B=1/1.2+1/2.3+...+1/49.50

ta có:

A=1/1^2+1/2^2+1/3^2+1/4^2+....+1/50^2<B=1/1.2+1/2.3+...+1/49.50 (1)

B=1/1.2+1/2.3+...+1/49.50

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

=1-1/50<1 (2)

từ (1) va (2)=>A<B<2

=>A<2

dễ lắm bn ạ  nhưng mk ko bt lm hhh