Question

Cho:

\(C=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2017}{4^{2017}}\)

\(CMR:C< \dfrac{1}{2}\)

đại ka tú giúp em với

Answer 1

\(C=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\)

\(\Rightarrow4C=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)

\(\Rightarrow3C=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)

\(\Rightarrow3C=1+\dfrac{1-\dfrac{1}{4^{2016}}}{3}-\dfrac{2017}{4^{2016}}\)

\(\Rightarrow C=\dfrac{1}{3}+\dfrac{1-\dfrac{1}{4^{2016}}}{9}-\dfrac{2017}{4^{2016}.3}< \dfrac{1}{2}\)

Vậy...

Answer 2

Ta có:

\(C=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\)

\(4C=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2017}{4^{2016}}\)

\(\Rightarrow3C=4C-C=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)Đặt \(D=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}\)

\(\Rightarrow4D=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2015}}\)

\(\Rightarrow3D=4D-D=4-\dfrac{1}{4^{2016}}\)

Ta có : Do \(4-\dfrac{1}{4^{2016}}< 4\Rightarrow3D< 4\)

\(\Rightarrow D< \dfrac{4}{3}\)

Thay giá trị tìm được của D vào 3C được:

\(3C=4-\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)

Do \(D< \dfrac{4}{3}\Rightarrow D-\dfrac{2017}{4^{2017}}< \dfrac{4}{3}\)

Hay \(3C< \dfrac{4}{3}\Rightarrow C< \dfrac{4}{9}< \dfrac{1}{2}\Rightarrow C< \dfrac{1}{2}\left(đpcm\right)\)

Vậy....

tik mik nha !!!

Answer 3

Mk thử giải lại xem có nhớ thôi chứ ko có ý j:v

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

\(4C=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2017}{4^{2016}}\)

\(4C-C=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2016}}\right)\)

\(3C=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)

Đặt:

\(\Rightarrow D=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}\)

\(4D=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)

\(4D=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)

\(4D-D=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3D=4-\dfrac{1}{4^{2016}}\)

\(D=\dfrac{4}{3}-\dfrac{1}{4^{2016}.3}\)

Thay vào ta có:
\(3C=\dfrac{4}{3}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2016}}\)

\(3C< \dfrac{4}{3}\Rightarrow C< \dfrac{4}{9}\)

\(\Rightarrow C< \dfrac{1}{2}\)