`Answer:`

\(T=\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)

\(\Leftrightarrow2T=2+\frac{3}{2}+\frac{4}{2^2}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\)

\(\Leftrightarrow2T-T=2+\left(\frac{3}{2}-\frac{2}{2}\right)+\left(\frac{4}{2^2}-\frac{4}{2^2}\right)+...+\left(\frac{2017}{2^{2015}}-\frac{2016}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

\(\Leftrightarrow2T-T=2+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

Ta đặt \(V=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(\Rightarrow T=2+V-\frac{2017}{2^{2016}}\text{(*)}\)

\(\Leftrightarrow2V=1+\frac{1}{2}+...+\frac{1}{2^{2014}}\)

\(\Leftrightarrow2V-V=\left(1+\frac{1}{2}+...+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)

\(\Leftrightarrow2V-V=1-\frac{1}{2^{2015}}\text{(**)}\)

Từ (*)(**)\(\Rightarrow T=2+\left(1-\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

\(\Leftrightarrow T=3-\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)

`=>T<3`

Ta có: \(T=\frac{2}{2^1}+\frac{3}{2^2}+\cdots+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)

=>2T=\(2+\frac32+\cdots+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\)

=>2T-T=\(2+\frac32+\cdots+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}-\frac{2}{2^1}-\frac{3}{2^2}-\cdots-\frac{2016}{2^{2015}}-\frac{2017}{2^{2016}}\)

=>T=\(2+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)

Đặt \(A=\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2015}}\)

=>2A=\(1+\frac12+...+\frac{1}{2^{2014}}\)

=>2A-A=\(1+\frac12+\cdots+\frac{1}{2^{2014}}-\frac12-\frac{1}{2^2}-\cdots-\frac{1}{2^{2015}}\)

=>\(A=1-\frac{1}{2^{2015}}=\frac{2^{2015}-1}{2^{2015}}\)

Ta có: \(T=2+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)

\(=2+\frac{2^{2015}-1}{2^{2015}}-\frac{2017}{2^{2016}}=2+\frac{2^{2016}-2-2017}{2^{2016}}=2+1-\frac{2019}{2^{2016}}<3\)

=>T<3

khó quá 

A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)

A=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)

A=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2015.2016}\right)\)

A=\(\frac{1}{4}-\frac{1}{2015.2016.2}\)\(\Rightarrow A<\frac{1}{4}\)

A<1

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