Question

[Nguyễn Như Anh]

Cho T= 2/2¹+3/2²+4/2³+...+2017/2²⁰¹⁶

So sánh T với 3

Answer 1

[Phùng Minh Quân]

Ta có : 

\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\)

\(\frac{1}{2}T=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\)

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

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

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

\(\frac{1}{2}T=1+\left(\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{2^{2016}}\)

\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

\(A=\frac{1}{2}-\frac{1}{2^{2016}}\)

Mà \(\frac{1}{2^{2016}}>0\)

\(\Rightarrow\)\(A=\frac{1}{2}-\frac{1}{2^{2016}}< \frac{1}{2}\)

\(\Leftrightarrow\)\(1+A-\frac{2017}{2^{2017}}< 1+\frac{1}{2}-\frac{1}{2^{2016}}-\frac{2017}{2^{2017}}\)

\(\Leftrightarrow\)\(\frac{1}{2}T< \frac{3}{2}-\left(\frac{1}{2^{2016}}+\frac{2017}{2^{2017}}\right)\)

Mà \(\frac{1}{2^{2016}}+\frac{2017}{2^{2017}}\)

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

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

\(\Rightarrow\)\(T< 3\)

Vậy \(T< 3\)

Chúc bạn học tốt ~ 

Answer 2

[Kang Nhầu]

\(T< 3\)