a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)(áp dụng quy tắc dấu ngoặt )
\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^8}\)
\(3A-A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^7}-\frac{1}{3^7}\right)-\frac{1}{3^8}\)
\(\Rightarrow2A=1+0+0...+0-\frac{1}{3^8}\)
\(2A=1-\frac{1}{3^8}\)
\(2A=\frac{3^8-1}{3^8}\)
\(A=\frac{3^8-1}{3^8}\div2=\frac{3^8-1}{3^8}.\frac{1}{2}=\frac{3^8-1}{3^8.2}\)
b) \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}\)
\(\Rightarrow B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)(áp dụng quy tắc dấu ngoặt )
\(B=\frac{1}{1}-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{100}-\frac{1}{100}\right)-\frac{1}{101}\)
\(B=\frac{1}{1}-0-0-0...-0-\frac{1}{101}\)
\(B=\frac{1}{1}-\frac{1}{101}\)
\(B=\frac{100}{101}\)
