Xét công thức tổng quát:
\(\frac{1}{n.\left(n+2\right)}=\frac{1}{2}.\frac{2}{n\left(n+2\right)}=\frac{1}{2}.\left(\frac{1}{n}-\frac{1}{n+2}\right)\\ \)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{60.62}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{60}-\frac{1}{62}\)
\(2A=\frac{1}{2}-\frac{1}{62}\)
\(2A=\frac{30}{62}\)
\(\Rightarrow A=\frac{30}{124}=\frac{15}{62}< \frac{15}{60}=\frac{1}{4}\)
Tương tự với câu 2 tính 6B
Ta có:
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{60.62}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{60.62}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{60}-\frac{1}{62}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{62}\right)=\frac{1}{2}.\frac{15}{31}=\frac{15}{62}\)
\(\Leftrightarrow\frac{15}{62}< \frac{1}{4}\Leftrightarrow A< \frac{1}{4}\left(dpcm\right)\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{201.206}\)
\(\Rightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{201.206}\right)\)
\(\Rightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{201}-\frac{1}{206}\right)\)
\(\Rightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{206}\right)=\frac{1}{5}.\frac{205}{206}=\frac{41}{206}\)
\(\Leftrightarrow\frac{41}{206}< \frac{1}{5}\Leftrightarrow B< \frac{1}{5}\left(dpcm\right)\)