Question

[Thủ lĩnh thẻ bài Sakura]

Cho A= 1 phần 1mũ 2+ 1 phần 2 mũ 2+...+1 phần 50 mũ 2 chứng minh A<2

Answer 1

[Kim Ngưu Ngốk]

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

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

Ta thấy \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3};......;\frac{1}{50^2}=\frac{1}{50.50}< \frac{1}{49.50}\)

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

\(B=1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)

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

Vì \(B< 2\)mà \(A< B\)nên \(A< 2\left(đpcm\right)\)

Answer 2

[Bùi Thế Hào]

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

 \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\); \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\);...; \(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)

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

=> \(A< 1+1-\frac{1}{50}\)

=> \(A< 2-\frac{1}{50}\)

=> \(A< 2\)