So sánh A với 1
A=1/1^2*2^2+5/2^2*3^2+7/3^2*4^2+...........+19/2015^2*2016*2
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A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
A=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)
A=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2015.2016}\right)\)
A=\(\frac{1}{4}-\frac{1}{2015.2016.2}\)\(\Rightarrow A<\frac{1}{4}\)
2A=2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ...+2/2014.2015.2016
Ta có: 2/1.2.3=1/1.2-1/2.3; 2/2.3.4=1/2.3-1/3.4; 2/3.4.5=1/3.4-1/4.5; ....; 2/2014.2015.2016=1/2014.2015-1/2015.2016
=> 2A=1/1.2-1/2015.2016
=> 2A < 1/2 => A < 1/4
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
A<1
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