Question

ChoA=1/5^2+1/6^2+...+1/100^2. Chứng minh rằng 1/6<A<1/4

Answer 1

*Có : 5² < 5.6 => \(\frac{1}{5^2}>\frac{1}{5.6}\)

         6² < 6.7 =>\(\frac{1}{6^2}>\frac{1}{6.7}\)

   ....

         100² < 100 . 101 => \(\frac{1}{100^2}>\frac{1}{100.101}\)

Cộng từng vế có :

\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...\frac{1}{100}-\frac{1}{101}\)

\(A>\frac{1}{5}-\frac{1}{101}\)

Mà \(\frac{1}{5}-\frac{1}{101}=\frac{101-5}{105}=\frac{96}{505}\)

=> \(A>\frac{96}{505}\)

Mà \(\frac{1}{6}=\frac{96}{576}< \frac{96}{505}\)

=> \(A>\frac{1}{6}\)(1)

*Có 5² > 5.4 => \(\frac{1}{5^2}< \frac{1}{5.4}\)

.......

    100² > 100.99 => \(\frac{1}{100^2}< \frac{1}{100.99}\)

Cộng từng vế có :

........ => A < \(\frac{96}{400}\)

Có \(\frac{1}{4}=\frac{100}{400}>\frac{96}{400}\)

=> A < \(\frac{1}{4}\)(2)

Từ (1)(2) => đpcm

Answer 2

\(\text{Ta thấy :}\)

\(\frac{1}{5^2}>\frac{1}{5.6}\)

\(\frac{1}{6^2}>\frac{1}{6.7}\)

\(......................................\)

\(\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(\Rightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A>\frac{1}{5}-\frac{1}{101}=\frac{101-5}{105}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)

\(\Rightarrow A>\frac{1}{6}\left(1\right)\)

\(\text{Lại thấy :}\)

\(\frac{1}{5^2}< \frac{1}{5.4}\)

\(\frac{1}{6^2}< \frac{1}{5.6}\)

\(..................................\)

\(\frac{1}{100^2}< \frac{1}{100.99}\)

\(\text{Tương tự như trên ta tính được }:\)

\(A< \frac{96}{400}< \frac{100}{400}=\frac{1}{4}\)

\(\Rightarrow A< \frac{1}{4}\left(2\right)\)

\(\text{Từ (1) và (2)}\Rightarrow\frac{1}{6}< A< \frac{1}{4}\)