Sửa đề: \(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+....+\dfrac{1}{2^{58}}\)
Ta có : \(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+.....+\dfrac{1}{2^{58}}\)
\(\Rightarrow2^3A=2^3.\left(\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+\dfrac{1}{2^{58}}\right)\)
\(\Rightarrow2^3A=1+\dfrac{1}{2}+\dfrac{1}{2^4}+.....+\dfrac{1}{2^{55}}\)
\(\Rightarrow2^3A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{55}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^7}+....+\dfrac{1}{2^{58}}\right)\)\(\Rightarrow7A=1-\dfrac{1}{2^{58}}\Rightarrow A=\dfrac{1-\dfrac{1}{2^{58}}}{7}\)
Vậy...........
~ Học tốt nha ~
\(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{58}}\\ 2^3\cdot A=\dfrac{2^3}{2}+\dfrac{2^3}{2^4}+\dfrac{2^3}{2^7}+...+\dfrac{2^3}{2^{58}}\\ 8A=4+\dfrac{1}{2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{55}}\\ 8A-A=\left(4+\dfrac{1}{2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{55}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{58}}\right)\\ 7A=4-\dfrac{1}{2^{58}}\\ A=\dfrac{4-\dfrac{1}{2^{58}}}{7}\)
