A= 2/1.4+2/4.7+2/7.10+...+2/97.100
= 2.(1/1.4+1/4.7+1/7.10+...+1/97.100)
= 2.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)
= 2.(1/1-1/100)
= 2.(99/100)
=99/50
\(\frac{2}{1.4}+\frac{2}{4.7}+....+\frac{2}{97.100}\)
\(=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{100}\right)\)
\(=\frac{1}{3}\left(2-\frac{2}{100}\right)=\frac{1}{3}\left(\frac{200}{100}-\frac{2}{100}\right)=\frac{1}{3}.\frac{198}{100}=\frac{33}{50}\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\frac{99}{100}\)
\(=\frac{33}{50}\)
Ta có:
\(A=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)
\(\Rightarrow\frac{3}{2}A=\frac{3}{2}\left(\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\right)\)
\(=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\Rightarrow\frac{3}{2}A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:\frac{3}{2}=\frac{33}{50}\)
Vậy \(A=\frac{33}{50}\)
21.4+24.7+....+297.10021.4+24.7+....+297.100
=13(21−24+24−27+...+297−2100)=13(21−24+24−27+...+297−2100)
=13(2−2100)=13(200100−2100)=13.198100=3350
