Tham khảo nha bạn :

Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến

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

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

a<2 ai k cho mik, mik se k lại hứa thế lun nói là làm

ta có:1/1^2=1/1

1/2^2=1/2*2<1/1*2=1/1-1/2

1/3^2=1/3*3<1/2*3=1/2-1/3

1/4^2=1/4*4<1/3*4

...

1/50^2=1/50*50<1/49*50=1/49-1/50

=>A=1/1-1/50+1

A=99/50<100/50=2

=>A<2

vậy A<2

\(\frac{1}{2^2}< \frac{1}{1.2}\)

...................\(\frac{1}{50^2}< \frac{1}{49.50}\)

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

\(\Rightarrow A< 1-\frac{1}{50}< \frac{49}{50}< 1< 2\)

1/2^2<1/1*2;1/3^2<1/2*3;1/4^2<1/3*4;1/50^2<1/49*50

ta có:

   =>    1/1^2+1/2*3+1/3*4+...+1/49*50

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

  <=>   1-1/50 < 2

    =>   A < 2

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

  =\(1+\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

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

                                                                                          \(< 1+1-\frac{1}{50}=\frac{99}{50}< 2\)

                     => \(A< 2\)

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

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

Ta thấy \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)

Khi đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{49.50}=B\)

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

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

Vì \(A< 1+B\)mà \(B< 1\)nên \(B+1< 2\)do đó \(A< 2\)

Vậy \(A< 2\)

1/1²+1/2²+....+1/50²<1/1+1/1x2+1/2x3+....+1/49x50=1-1/50=49/50<2

=>A<2(đpcm)

Ta co 

1/2^2<1/1-1/2

1/3^2<1/2-1/3

1/4^2<1/3-1/4

...

1/50^2<1/49-1/50

=>1/1^2+...+1/50^2<1/1-1/2+1/2-1/3+...+1/49-1/50=1/1-1/50=49/50

Ma 49/50<2

=> 1/1^2+1/2^2+...+1/50^2<2

\(A<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1+\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

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

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

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

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

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

=>A<2  (đpcm)

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

\(A=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)=1+B\)( B là biểu thức trong ngoặc )

Xét B

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

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

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

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

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

Vậy B < 1

\(\Rightarrow A=1+B<1+1=2\)

\(\Rightarrow A<2\)

Ta có: 1/2^2 < 1/1.2

          1/3^2 < 1/2.3

          1/4^2 < 3.4

            ............

          1/50^2 < 1/49.50

=>  A= 1 + 1/2^2 + 1/3^2 + ....+1/50^2 < 1 + 1/1.2 + 1/2.3 +....+1/49.50 = 1+ 1-1/2+ 1/2 - 1/3 +. ... + 1/49 -1/50 = 1 + 1 - 1/50 = 2-1/50 < 2

Vậy A < 2 (ĐPCM)