Chủ đề này có 1 trả lời

[thanhmylam]

Chứng minh $\frac{3}{2^{3}}+\frac{4}{2^{4}}+...+\frac{2014}{2^{2014}}+\frac{2015}{2^{2015}}< 1$

[huykinhcan99]

Chứng minh $\frac{3}{2^{3}}+\frac{4}{2^{4}}+...+\frac{2014}{2^{2014}}+\frac{2015}{2^{2015}}< 1$

 

Đặt $S=\dfrac{3}{2^{3}}+\dfrac{4}{2^{4}}+\ldots+\dfrac{2014}{2^{2014}}+\dfrac{2015}{2^{2015}}$.

 

Ta có:

\begin{align*} S=2S-S&=2\left(\dfrac{3}{2^{3}}+\dfrac{4}{2^{4}}+\ldots+\dfrac{2014}{2^{2014}}+\dfrac{2015}{2^{2015}}\right)-\left(\dfrac{3}{2^{3}}+\dfrac{4}{2^{4}}+\ldots+\dfrac{2014}{2^{2014}}+\dfrac{2015}{2^{2015}}\right) \\ &=\dfrac{3}{2^{2}}+\left(\dfrac{4}{2^{3}}+\ldots+\dfrac{2014}{2^{2013}}+\dfrac{2015}{2^{2014}}\right)-\left(\dfrac{3}{2^{3}}+\dfrac{4}{2^{4}}+\ldots+\dfrac{2014}{2^{2014}}\right)-\dfrac{2015}{2^{2015}} \\ &= \dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\ldots+\dfrac{1}{2^{2014}}-\dfrac{2015}{2^{2015}} \\& = \left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\ldots+\dfrac{1}{2^{2014}}\right)-\dfrac{2015}{2^{2015}} \end{align*}

 

Đặt $A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\ldots+\dfrac{1}{2^{2014}}$. Ta lại có

\begin{align*} A=2A-A&=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\ldots+\dfrac{1}{2^{2013}}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\ldots+\dfrac{1}{2^{2014}}\right) \\ &= 1-\dfrac{1}{2^{2014}} \end{align*}

 

Khi đó $S=1-\dfrac{1}{2^{2014}}-\dfrac{2015}{2^{2015}}<1$.