Question

[Vũ Thành Dương]

\(ChoA=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)chứng minh rằng A<2

Answer 1

[Thắng Nguyễn]

đặt B=1/2.3+1/3.4+...+1/49.50

=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<1 (1)

Mà 1<2(2)

A =1/1+1/2.2+1/3.3+...+1/50.50<1-1/2+1/2-1/3+...+1/49-1/50 (3)

từ (1),(2),(3) =>A<2

Answer 2

[Bùi Minh Anh]

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

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

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

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

Vậy A nhỏ hơn 2