PH
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
8 tháng 3 2018
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 \(T< 3\)
Bạn xem đúng không nhé, chúc bạn học tốt ~
8 tháng 3 2018
bạn tham khảo tạm ở đây nhé
https://olm.vn/hoi-dap/question/994432.html
^^
8 tháng 3 2018
bạn tham khảo tại đây nhé
http://olm.vn/hoi-dap/question/994432.html
^-^
AQ
0
K
1
26 tháng 9 2016
Ta có:
\(\left(2015^{2015}+2016^{2015}\right)^{2016}=\left(2015^{2015}+2016^{2015}\right)^{2015}.\left(2015^{2015}+2016^{2015}\right)\)
\(>\left(2015^{2015}+2016^{2015}\right)^{2015}.2016^{2015}=\left[\left(2015^{2015}+2016^{2015}\right)2016\right]^{2015}\)
\(>\left(2015^{2015}.2015+2016^{2015}.2016\right)^{2015}=\left(2015^{2016}+2016^{2016}\right)^{2015}\)
Vậy \(\left(2015^{2015}+2016^{2015}\right)^{2016}>\left(2015^{2016}+2016^{2016}\right)^{2015}\)
CH
Cô Hoàng Huyền
Admin
VIP
23 tháng 9 2016
1. Ta sẽ chứng minh \(2015^{2016}>2016^{2015}\)
\(\Leftrightarrow2016^{2015}-2015^{2016}< 0\Leftrightarrow2016^{2016}-2016.2015^{2016}< 0\)
\(\Leftrightarrow2016.2016^{2016}-2015.2016^{2016}-2016.2015^{2016}< 0\)
\(\Leftrightarrow2016\left(2016^{2016}-2015^{2016}\right)< 2015.2016^{2016}\)
\(\Leftrightarrow2016\left(2016^{2015}+2016^{2014}.2015+...+2015^{2015}\right)< 2015.2016^{2016}\)
\(\Leftrightarrow2016^{2015}.2015+...+2016.2015^{2015}< 2014.2016^{2016}\)
\(\Leftrightarrow2016^{2014}.2015+2016^{2013}.2015^2+...+2015^{2015}< 2014.2016^{2015}\)
\(\Leftrightarrow2015^{2015}< \left(2016^{2015}-2015.2016^{2014}\right)+\left(2016^{2015}-2015^2.2016^{2013}\right)\)
\(+...+\left(2016^{2015}-2015^{2014}.2016\right)\)
\(\Leftrightarrow2015^{2015}< 2014.2016^{2014}+2013.2016^{2014}.2015+...+2016.2015^{2013}\)
Lại có \(2015^{2015}=2014.2015^{2014}+2015^{2014}< 2014.2016^{2014}+2015^{2014}\)
Mà \(2015^{2014}< 2013.2016^{2014}.2015\)
nên \(2015^{2014}< 2014.2016^{2014}+2013.2016^{2014}.2015+...+2016.2015^{2013}\)
Vậy \(2015^{2016}>2016^{2015}.\)