xét T=2/2+3/22+4/23+...+2015/22014. Hãy so sánh T với 3
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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 ~
PH
0
LK
2
27 tháng 6 2021
Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)
\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)
K
1
NT
0
YN
4 tháng 4 2022
`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`
DB
0