Question

a) Tính S = \(\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)

b) Cho A = \(\dfrac{1}{2017}\)+ \(\dfrac{2}{2017^2}\) + \(\dfrac{3}{2017^3}\) + ... + \(\dfrac{2017}{2017^{2017}}\) + \(\dfrac{2018}{2017^{2018}}\)

Chứng minh tằng A < \(\dfrac{2017}{2016^2}\)

Nhanh lên nha chiều mình học rồi

Answer 1

a) \(S=\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)

\(\Rightarrow2S=\dfrac{2\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)

\(2S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}\)

\(\Rightarrow2S-S=S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}-\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)

\(S=\dfrac{\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)

\(S=\dfrac{2^{2018}-2}{1-2^{2017}}=\dfrac{-2\left(1-2^{2017}\right)}{1-2^{2017}}=-2\) vậy \(S=-2\)