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`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
Ta có :
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)
\(T=1+\frac{3}{1.2^2}+\frac{4}{2.2^2}+\frac{5}{2^2.2^2}+...+\frac{2016}{2^{2013}.2^2}+\frac{2017}{2^{1014}.2^2}\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{4}+\frac{6}{8}+...+\frac{2016}{x}+\frac{2017}{x}\right)\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2013}}+\frac{2017}{2^{2014}}\right)\)
Đến chỗ này chịu!
Ta có
\(T=1+\frac{3}{1\cdot2^2}+\frac{4}{2\cdot2^2}+...+\frac{2017}{2^2\cdot2^{2014}}\)
\(T=1+\frac{1}{2^2}\cdot\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\right)\)
T=1+ 1.2 2 3 + 2.2 2 4 + 2 2 .2 2 5 +...+ 2 2013 .2 2 2016 + 2 1014 .2 2 2017 = 1 + 1 2 2 . ( 3 + 2 + 5 4 + 6 8 + . . . + 2016 x + 2017 x ) =1+ 2 2 1 .(3+2+ 4 5 + 8 6 +...+ x 2016 + x 2017 ) = 1 + 1 2 2 . ( 3 + 2 + 5 2 2 + 6 2 3 + . . . + 2016 2 2013 + 2017 2 2014 ) =1+ 2 2 1 .(3+2+ 2 2 5 + 2 3 6 +...+ 2 2013 2016 + 2 2014 2017 ) Đến chỗ này chịu!