Question

1/ Cho \(A=\frac{1}{1.102}+\frac{1}{2.203}+...+\frac{1}{299.400}\)

Chứng minh rằng: \(A=\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)

2/ Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\)

           \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)

a) So sánh \(A\)với \(B\)                                                                   b) Chứng minh: \(A< 1\)

3/ Cho \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

Chứng minh: \(\frac{1}{2}< A< 1\)

                             GIÚP MÌNH NHA, MÌNH CẢM ƠN. MÌNH ĐANG CẦN GẤP!!!!

Answer 1

2/

a, Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2005^2}< \frac{1}{2004.2005}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}=B\)

b, \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{2005}=1-\frac{1}{2005}< 1\)

3/ 

Ta có: \(\frac{1}{101}< \frac{1}{100};\frac{1}{102}< \frac{1}{100};...;\frac{1}{200}< \frac{1}{100}\)

\(\Rightarrow A< \frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\left(100ps\right)=\frac{1}{100}\cdot100=1\left(1\right)\)

Lại có: \(\frac{1}{101}>\frac{1}{200};\frac{1}{102}>\frac{1}{200};...;\frac{1}{200}=\frac{1}{200}\)

\(\Rightarrow A>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\left(100ps\right)=\frac{1}{200}\cdot100=\frac{1}{2}\left(2\right)\)

Từ (1) và (2) => \(\frac{1}{2}< A< 1\)

Answer 2

vận dụng 3 A nha bn

dễ mà

xong tìm A ok nha bn

ok

Answer 3

1/

\(A=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)

=\(\frac{1}{101}\left(\frac{101}{1.102}+\frac{101}{2.103}+...+\frac{101}{299.400}\right)\)

=\(\frac{1}{101}\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)

=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+...+\frac{1}{400}\right)\right]\)

=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)+\left(\frac{1}{102}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+...+\frac{1}{299}\right)-\left(\frac{1}{300}+...+\frac{1}{401}\right)\right]\)

=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+...+\frac{1}{401}\right)\right]\)