Question

a,Cho A=\(\frac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)  

Tinh A.

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

CMR: A<\(\frac{2017}{2017^2}\)

Mik dang can gap.Giup mik voi.Thanks nhiu^^

Answer 1

a) \(A=\frac{2+2^2+...+2^{2017}}{1-2^{2017}}\)

Đặt \(B=2+2^2+...+2^{2017}\)

\(\Rightarrow2B=2^2+2^3+...+2^{2018}\)

\(\Rightarrow2B-B=\left(2^2+2^3+...+2^{2018}\right)-\left(2+...+2^{2017}\right)\)

\(\Rightarrow B=2^{2018}-2\)

\(\Rightarrow A=\frac{2^{2018}-2}{1-2^{2017}}\)

\(\Rightarrow A=\frac{-2.\left(1-2^{2017}\right)}{1-2^{2017}}\)

\(\Rightarrow A=-2\)

Answer 2

b)Đề phải là CM: \(A< \frac{2017}{2016^2}\)

 \(A=\frac{1}{2017}+\frac{2}{2017^2}+...+\frac{22017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)

\(\Rightarrow2017A=1+\frac{2}{2017}+...+\frac{22017}{2017^{2016}}+\frac{2018}{2017^{2017}}\)

\(\Rightarrow2017A-A=\left(1+...+\frac{2018}{2017^{2017}}\right)-\left(\frac{1}{2017}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\right)\)

\(\Rightarrow2016A=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}-\frac{2018}{2017^{2018}}\)

Đặt \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)

\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)

\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)

\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)

\(\Rightarrow2016S< 2017\)

\(\Rightarrow S< \frac{2017}{2016}\)

\(\Rightarrow2016A< \frac{2017}{2016}\)

\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)