Question

[Nguyễn Tiến Đạt]

Xét tổng T= \(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2015}{2^{2014}}\).Hãy so sánh  T với 3

Answer 1

[Phùng Minh Quân]

Ta có : 

\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2015}{2^{2014}}\)

\(\frac{1}{2}T=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2015}{2^{2015}}\)

\(T-\frac{1}{2}T=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2015}{2^{2014}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2015}{2^{2015}}\right)\)

\(\frac{1}{2}T=1+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2015}{2^{2014}}-\frac{2}{2^2}-\frac{3}{2^3}-\frac{4}{2^4}-...-\frac{2015}{2^{2015}}\)

\(\frac{1}{2}T=1+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2015}{2^{2014}}-\frac{2014}{2^{2014}}\right)-\frac{2015}{2^{2015}}\)

\(\frac{1}{2}T=1+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\right)-\frac{2015}{2^{2015}}\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\)

\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)

\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\right)\)

\(A=\frac{1}{2}-\frac{1}{2^{2014}}\)

Mà \(\frac{1}{2^{2014}}>0\)

\(\Rightarrow\)\(A=\frac{1}{2}-\frac{1}{2^{2014}}< \frac{1}{2}\)

\(\Leftrightarrow\)\(1+A-\frac{2015}{2^{2015}}< 1+\frac{1}{2}-\frac{1}{2^{2014}}-\frac{2015}{2^{2015}}\)

\(\Leftrightarrow\)\(\frac{1}{2}T< \frac{3}{2}-\left(\frac{1}{2^{2014}}+\frac{2015}{2^{2015}}\right)\)

Mà \(\frac{1}{2^{2014}}+\frac{2015}{2^{2015}}>0\)

\(\Rightarrow\)\(\frac{1}{2}T< \frac{3}{2}\)

\(\Rightarrow\)\(\frac{1}{2}T.2< \frac{3}{2}.2\)

\(\Rightarrow\)\(T< 3\) ( đpcm ) 

Vậy \(T< 3\)

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Answer 2

[Nguyễn Tiến Đạt]

thank