A=\(1+2^1+2^2+2^3+2^4+...+2^{2015}\)
2A=\(2^1+2^2+2^3+2^4+...+2^{2015}+2^{2016}\)
2A-A=\(2^{2016}-1\)
Vậy A=\(2^{2016}-1\)
\(A=1+2^1+2^2+2^3+....+2^{2015}\)
\(2A=2\left(1+2+2^2+.....+2^{2015}\right)\)
\(2A=2+2^2+2^3+....+2^{2016}\)
\(2A-A=\left(2+2^2+.....+2^{2016}\right)-1-2-2^2-...-2^{2015}\)
\(A=2^{2016}-1\)
Vậy\(A=---\)
Cách làm của mình gần giống bạn Đoàn Khánh Linh nhưng mình sẽ viết cho ngắn gọn:
Ta có:
\(A=1+2^1+2^2+2^3+........+2^{2015}\)
\(2A=2+2^2+2^3+2^4+.........+2^{2016}\)
\(2A-A=\left(2+2^2+2^3+2^4+.......+2^{2016}\right)-\left(1+2^1+2^2+2^3+.........+2^{2015}\right)\)
\(A=2^{2016}-1\)
Vậy A = \(2^{2016}-1\)
