Question

Bài 1. Chứng minh:

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

Answer 1

Ta có:

$\backslash (A=\; \backslash frac\{1\}\{2^2\}+\backslash frac\{1\}\{3^2\}+...+\backslash frac\{1\}\{100^2\}<\; \backslash frac\{1\}\{1.2\}+\backslash frac\{1\}\{2.3\}+\backslash frac\{1\}\{3.4\}+...+\backslash frac\{1\}\{99.100\}\backslash )$

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

\(=\frac{99}{100}< \frac{100}{100}=1\)

Vậy: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\)

Answer 2

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

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 1\)

Vậy \(A< 1\) (Đpcm)

Answer 3

Làm lại nha cái kia bị lỗi mất rồi .-.

\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< \frac{100}{100}\)

\(=1\)

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