Question

[nguyễn thu ánh]

CMR:1 phần 2^2+ 1 phần 3^2+ 1 phần 4^2+..+ 1 phần 100^2<3 phần4

Answer 1

[Ashes PK249]

\(A=\frac{1}{5}+\frac{1}{15}+...+\frac{1}{10000}\)

\(5A=1+\frac{1}{5}+...+\frac{1}{2000}\)

\(\rightarrow4A=1-\frac{1}{10000}\leftrightarrow A=\frac{1-\frac{1}{10000}}{4}\) TA CÓ: \(1-\frac{1}{10000}< 1< 3\)\(\rightarrow A< \frac{3}{4}\rightarrowĐPCM\)

Answer 2

[Akai Haruma]

Lời giải:

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

\(2A=\frac{2}{2^2}+\frac{2}{3^2}+....+\frac{2}{100^2}\)\(<\underbrace{ \frac{2}{2^2-1}+\frac{2}{3^2-1}+\frac{2}{4^2-1}+....+\frac{2}{100^2-1}}_{M}\)

Mà:

\(M=\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+\frac{2}{4.6}+....+\frac{2}{99.101}\)

\(=\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)+\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

\(=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-....+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\left(1-\frac{1}{101}\right)+\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{3}{2}-\frac{1}{101}-\frac{1}{100}< \frac{3}{2}\)

Do đó: $2A< \frac{3}{2}\Rightarrow A< \frac{3}{4}$