Ta có : 

\(\left(\sqrt{2015}+\sqrt{2017}\right)^2=2015+2\sqrt{2015.2017}+2017=8064+2\sqrt{2015.2017}\)

\(\left(2\sqrt{2016}\right)^2=8064\)

Vì \(\left(\sqrt{2015}+\sqrt{2017}\right)^2>\left(2\sqrt{2016}\right)^2\) nên \(\sqrt{2015}+\sqrt{2017}>2\sqrt{2016}\)

Vậy... 

Chúc bạn học tốt ~ 

Cảm ơn bn nhiều nhé :)))

A=(1+2+2^2)+2^3(1+2+2^2)+...+2^2013(1+2+2^2)+2^2016

=7(1+2^3+...+2^2013)+2^2016

Vì 2^2016 chia 7 dư 1

nên A chia 7 dư 1

Ta có: \(A=1+2+2^2+...+2^{2015}\)

\(2A=2\cdot\left(1+2+2^2+...+2^{2015}\right)\)

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

\(2A-A=2+2^2+...+2^{2016}-1-2-2^2-...-2^{2015}\)

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

A không thể biết dưới dạng lũy thừa của 8 được 

A=22016−1

`#3107`

\(A=1+2^1+2^2+2^3+...+2^{2015}\)

\(2A=2+2^2+2^3+2^4+...+2^{2016}\)

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

\(A=2+2^2+2^3+2^4+...+2^{2016}-1-2-2^2-2^3-...-2^{2015}\)

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

Vậy, \(A=2^{2016}-1.\)

\(A=2^0+2^1+2^2+...+2^{2015}\)

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

\(A=2A-A=2^{2016}-2^0\)

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

`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 : 

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

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

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

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

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

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

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 ~ 

\(T< 3\)