đặt \(A=1+2+2^2+2^3+...............+2^{2016}\)
\(2A=2+2^2+2^3+2^4+....................+2^{2017}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2017}\right)-\left(1+2+2^2+2^3+...+2016\right)\)
\(A=2^{2017}-1\)
thay vào S, ta được:
\(S=\dfrac{2^{2017}-1}{1-2^{2017}}=-1\)