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mình gợi ý nè : bạn thử lấy T nhân với 2 xem ( cả hai vế nhé )
Nếu bạn không ra thì k cho mình đi mình trình bày cho đôn giản mà mỗi tội hơi dài một chút.
`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`
1/2T=2/22 +3/23 +4/24 +...+2017/22017
T-1/2T= (2/21+3/22+4/23+...+2017/22016)-(2/22+3/23+4/24+...+2017/22017)
1/2T=2/21+3/22+4/23+...+2017/22016-2/22-3/23-4/24-...-2017/22017
1/2T=1+(3/22-2/22)+(4/23-3/23)+...+(2017/22016-2016/22016)-2017/22017
1/2T=1+(1/22+1/23+1/24+...+1/22016)-2017/22017
xét A = 1/22+1/23+1/24+...+1/22016
phần này dễ bạn tự làm nhé
A=1/2-1/22016<1/2(vì 1/22016>0)
1/2T<1/21+1/2-(1/22016+2017/22017)
1/2T<3/2(vì 1/22016+2017/22017>0)
T<3/2:1/2
T<3
vậy T<3
1/2T=2/22 +3/23 +4/24 +...+2017/22017 T-1/2T= (2/21+3/22+4/23+...+2017/22016 )-(2/22+3/23+4/24+...+2017/22017 ) 1/2T=2/21+3/22+4/23+...+2017/22016 -2/22 -3/23-4/24 -...-2017/22017 1/2T=1+(3/22 -2/22 )+(4/23 -3/23 )+...+(2017/22016 -2016/22016 )-2017/22017 1/2T=1+(1/22+1/23+1/24+...+1/22016 )-2017/22017 xét A = 1/22+1/23+1/24+...+1/22016 phần này dễ bạn tự làm nhé A=1/2-1/22016<1/2(vì 1/22016>0) 1/2T<1/21+1/2-(1/22016+2017/22017 ) 1/2T<3/2(vì 1/22016+2017/22017>0) T<3/2:1/2 T<3
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\) => \(\frac{T}{2}=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\)
=> \(T-\frac{T}{2}=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\right)\)
<=> \(\frac{T}{2}=\frac{2}{2^1}+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2017}{2^{2016}}-\frac{2016}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
<=> \(\frac{T}{2}=1+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
Đặt: \(M=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}=>2M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\)
=> \(2M-M=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
=> \(M=\frac{1}{2}-\frac{1}{2^{2016}}< \frac{1}{2}\)
=> \(\frac{T}{2}< 1+\frac{1}{2}-\frac{2017}{2^{2017}}< 1+\frac{1}{2}=\frac{3}{2}\)
=> T < 3
T < 3