Question

I. Tính các tổng sau :

\(A=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)

\(B=\frac{12}{84}+\frac{12}{210}+\frac{12}{390}+...+\frac{12}{2100}\)

GIÚP MÌNH VỚI!

Answer 1

\(A=2\left(\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{380}\right)=2\left(\frac{6-5}{5.6}+\frac{7-6}{6.7}+.....+\frac{20-19}{20.19}\right)=2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)=2\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{3}{10}\)

\(B=\frac{12}{84}+\frac{12}{210}+.....+\frac{12}{2100}=\frac{4}{28}+\frac{4}{70}+.....+\frac{4}{700}=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{1}{25.28}\right)=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.....-\frac{1}{28}\right)=\frac{4}{2}.\frac{6}{28}=\frac{3}{7}\)