*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

\(\text{Ta thấy :}\)

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

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

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

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

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

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

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

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

\(\text{Lại thấy :}\)

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

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

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

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

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

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

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

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

Chung minh rang abcabcchia het cho 37

tra loi giup minh cau nay voi

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}< 1< 2\Rightarrow A< 2\Rightarrowđpcm\)

thanks ban vi minh dang rat can dap an nay

Ta có \(A

đặt B=1/2.3+1/3.4+...+1/49.50

=1/1.2+1/2.3+1/3.4+...+1/49.50

=1-1/2+1/2-1/3+...+1/49-1/50

=1-1/50<1 (1)

Mà 1<2(2)

A =1/1+1/2.2+1/3.3+...+1/50.50<1-1/2+1/2-1/3+...+1/49-1/50 (3)

từ (1),(2),(3) =>A<2

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

=> \(A<1+\frac{1}{1.2}+\frac{1}{2.3}+.............+\frac{1}{49.50}\)

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

=> \(A<2-\frac{1}{50}\Rightarrow A<2\)

Vậy A nhỏ hơn 2

\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{27}+2^{28}+2^{29}\right)\\ A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{27}\left(1+2+2^2\right)\\ A=\left(1+2+2^2\right)\left(1+2^3+...+2^{27}\right)\\ A=7\left(1+2^3+...+2^{27}\right)⋮7\)