A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7A=(1+\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}})-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6A=\left(1-\frac{1}{7^{99}}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{7^{99}}\right):6\)
Câu b tương tự nha
a) \(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...........+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.........+\frac{1}{7^{99}}\)
\(\Rightarrow7A-A=6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\frac{A}{7}=\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\)
\(A-\frac{A}{7}=\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\right)\)
\(\frac{6}{7}A=\frac{1}{7}-\frac{1}{7^{101}}\)
\(A=\left(\frac{1}{7}-\frac{1}{7^{101}}\right).\frac{7}{6}\)
\(A=\frac{1}{6}-\frac{1}{6.7^{100}}\)
\(B=\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
\(=4.\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)\)
Gọi \(C=\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\)
\(\frac{C}{5}=\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\)
\(C-\frac{5}{C}=\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)-\left(\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\right)\)
\(\frac{4}{5}C=\frac{1}{5}-\frac{1}{5^{201}}\)
\(C=\left(\frac{1}{5}-\frac{1}{5^{201}}\right).\frac{5}{4}\)
\(=\frac{1}{4}-\frac{1}{4.5^{200}}\)
Thay vào B ta có
\(B=4.\left(\frac{1}{4}-\frac{1}{4.5^{200}}\right)\)
=\(=1-\frac{1}{5^{200}}\)
