Tổng này không nên tính giá trị cụ thể bạn nhé.

c) P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

Dễ thấy \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}.50=\dfrac{1}{3}\)(1)

Tương tự

 \(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>50.\dfrac{1}{200}=\dfrac{1}{4}\)(2) 

Từ (1) và (2) ta được

\(P>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) 

P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

         \(\overline{50\text{ hạng tử }}\)                            \(\overline{50\text{ hạng tử }}\)

\(< \left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)+\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)\) 

\(=\dfrac{1}{100}.50+\dfrac{1}{150}.50=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\Rightarrow P< \dfrac{5}{6}< 1\)

\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+...+\dfrac{1}{101.400}\)

\(\Rightarrow299A=\dfrac{299}{1.300}+\dfrac{299}{2.301}+...+\dfrac{299}{101.400}=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+...+\dfrac{1}{101}-\dfrac{1}{400}=M\)

\(\Rightarrow A=\dfrac{M}{299}\left(1\right)\)

Ta lại có:

\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+...+\dfrac{1}{298.399}+\dfrac{1}{299.400}\)

\(\Rightarrow101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+...+\dfrac{101}{399.400}=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+...+\dfrac{1}{399}-\dfrac{1}{400}=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+...+\dfrac{1}{101}-\dfrac{1}{400}=M\)

\(\Rightarrow B=\dfrac{M}{101}\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{M}{299}:\dfrac{M}{101}=\dfrac{101}{299}\)

a )   Số lượng số của dãy số trên là : 

\(\left(200-101\right):1+1=100\) ( số ) 

Do \(100⋮2\)nên ta nhóm dãy số trên thành 2 nhóm như sau : 

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)\)

\(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};...;\frac{1}{149}>\frac{1}{150};\frac{1}{150}=\frac{1}{150}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}.50=\frac{1}{3}\left(1\right)\)

\(\frac{1}{151}>\frac{1}{200};\frac{1}{152}>\frac{1}{200};...;\frac{1}{199}>\frac{1}{200};\frac{1}{200}=\frac{1}{200}\)

\(\Rightarrow\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}>\frac{1}{200}.50=\frac{1}{4}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{3}+\frac{1}{4}=\frac{7}{2}\left(3\right)\)

\(\frac{1}{101}< \frac{1}{100};\frac{1}{102}< \frac{1}{100};...;\frac{1}{199}< \frac{1}{100};\frac{1}{200}< \frac{1}{100}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}< \frac{1}{100}.100=1\left(4\right)\)

Từ \(\left(3\right);\left(4\right)\Rightarrowđpcm\)

b )  Số lượng số dãy số trên là : 

\(\left(150-101\right):1+1=50\)( số ) 

Ta có : \(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};\frac{1}{103}>\frac{1}{150};...;\frac{1}{150}=\frac{1}{150}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}.50=\frac{1}{3}\)

\(\Rightarrowđpcm\)

\(3A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{103}}-\dfrac{1}{3^{104}}\)

=>\(4A=3-\dfrac{1}{3^{104}}=\dfrac{3^{105}-1}{3^{104}}\)

=>\(A=\dfrac{3^{105}-1}{3^{104}\cdot4}\)

\(3A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{103}}-\dfrac{1}{3^{104}}\)

=>\(4A=3-\dfrac{1}{3^{104}}=\dfrac{3^{105}-1}{3^{104}}\)

=>\(A=\dfrac{3^{105}-1}{3^{104}\cdot4}\)