Question

[Lưu Chúc Loan]

Chứng tỏ: A =1/3 mũ 2+ 1/4 mũ 2+ 1/5 mũ 2+... +1/100 <1/2

Answer 1

[❤❤❤Huyền .... Trân ❤❤❤]

A=\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+.........+\(\frac{1}{100^2}\)

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

 \(\frac{1}{4^2}\)<\(\frac{1}{3.4}\)

\(\Rightarrow\)\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)<\(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

=>\(\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)< \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)

=> \(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+.....+\(\frac{1}{100^2}\)< \(\frac{1}{2}-\frac{1}{100}\)

=>A< \(\frac{1}{2}\)

Answer 2

[Phạm Trà Giang]

Ta có: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)

Ta thấy: \(\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};\frac{1}{5^2}< \frac{1}{4\cdot5}...\frac{1}{100^2}< \frac{1}{99\cdot100}\)

\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{100}\Rightarrow A< \frac{1}{2}\left(ĐPCM\right)\)