Question

[Nguyễn Mỹ Hạnh]

chứng minh A<2

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

Answer 1

[Hoàng Phúc]

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)

Answer 2

[Nguyễn Xuân Sáng]

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

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

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

\(\Rightarrow A<2\)