Tính:
\(A=\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2017}{4^{2017}}\)
\(A=\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2017}{4^{2017}}\)
\(4A=4\left(\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2017}{4^{2017}}\right)\)
\(4A=\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2017}{4^{2016}}\)
\(4A-A=\left(\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=\dfrac{2}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)
Đặt:
\(G=\dfrac{2}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)
\(4G=4\left(\dfrac{2}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)
\(4G=2+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{4}{2^{2015}}\)
\(4G-G=\left(2+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{4}{2^{2015}}\right)-\left(\dfrac{2}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3G=\dfrac{9}{4}-\dfrac{2}{4}-\dfrac{1}{4^{2016}}\)
\(3G=\dfrac{7}{4}-\dfrac{1}{4^{2016}}\)
\(G=\dfrac{7}{12}-\dfrac{1}{4^{2016}.3}\)
\(A=\dfrac{7}{12}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
Bổ sung từ thêm cho Phúc, Phúc sai đoạn cuối!
\(3G=\dfrac{7}{4}-\dfrac{1}{4^{2016}}\)
\(\Rightarrow G=\dfrac{7}{12}-\dfrac{1}{3.4^{2016}}\)
mà \(G=3A\) nên
\(\Rightarrow A=\dfrac{7}{36}-\dfrac{1}{9.4^{2016}}=\dfrac{7}{36}-\dfrac{4^{-2016}}{9}\)
Vậy..............
Chúc bạn học tốt!!!
