Question

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

Answer 1

Đặt A=1/1²+1/2²+1/3²+1/4²+....+1/50²<1/1.2+1/2.3+1/3.4+....+1/49.50

      A=1-1/50=50/50-1/50=49/50<2

      A=49/50<2 hay 98/100<100/100

Vậy A<2

Answer 2

Ta có

A= 1/2 + 1/2² + 1/3² + 1/4² +.......+ 1/50²

  =1/2.2 + 1/3.3 + 1/4.4 +.......+ 1/50.50

  <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 <100/50=2

Vậy A < 

Answer 3

Ta có : \(\frac{1}{2^2}<\frac{1}{1.2}\)

           \(\frac{1}{3^2}<\frac{1}{2.3}\)

          \(\frac{1}{4^2}<\frac{1}{3.4}\)

          ...................  

          ...................

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

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

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

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

Answer 4

hướng giả đúng rồi nhưng các b hơi nhầm

\(\frac{1}{50^2}<\frac{1}{50.51}\)

chứ mẫu k phải là 49.50