\(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{122.125}\)
\(3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{122.125}\)
Nhận xét:
\(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5}=\frac{3}{10}\)
\(\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8}=\frac{3}{40}\)
\(\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11}=\frac{3}{88}\)
=.
Từ nhận xét trên ta có:
\(3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{122}-\frac{1}{125}\)
\(3B=\frac{1}{2}-\frac{1}{125}=\frac{123}{250}\)
\(B=\frac{123}{250}:3=\frac{41}{250}\)
\(C=\frac{1}{9.11}+\frac{1}{11.13}+...+\frac{1}{97.99}\)
\(2C=\frac{2}{9.11}+\frac{2}{11.13}+...+\frac{2}{97.99}\)
Nhận xét:
\(\frac{2}{9.11}=\frac{1}{9}-\frac{1}{11}=\frac{2}{99}\)
\(\frac{2}{11.13}=\frac{1}{11}-\frac{1}{13}=\frac{1}{143}\)
=-..
Từ nhận xét trên ta có:
\(2C=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{99}\)
\(2C=\frac{1}{9}-\frac{1}{99}=\frac{10}{99}\)
\(C=\frac{10}{99}:2=\frac{5}{99}\)
