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

\(T=1+\frac{3}{1\cdot2^2}+\frac{4}{2\cdot2^2}+...+\frac{2017}{2^2\cdot2^{2014}}\) 

\(T=1+\frac{1}{2^2}\cdot\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\right)\)

M = 1 + 1/2 + 1/2² + ... + 1/2²⁰¹⁵ + 1/2²⁰¹⁶

=> 2.M = 2 + 1 + 1/2 + ... + 1/2²⁰¹⁴ + 1/2²⁰¹⁵

=> 2.M = 3 + 1/2 +...+ 1/2²⁰¹⁴ + 1/2²⁰¹⁵

=> 2.M - M = 3 -1 + 1/2²⁰¹⁶

=> M = 2 + 1/2²⁰¹⁶ 

=> M > 2 ( Do 2 + 1/2²⁰¹⁶  > 2 )

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(2015/2016)2016 lon hon(2016/2015)2016

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