\(5S=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2015}{5^{2014}}\Rightarrow4S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}-\frac{2015}{5^{2015}}\)

Đặt B = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}\)

 => 5B = \(5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)

=> 4B = \(5-\frac{1}{5^{2014}}

2A= 2+2^3+2^4+...+2^2015+2^2016

2A-A=2^2016-1

A=(2^2016-1):2

Vì (2^2016-1):2 bé hơn 2^2016 nên A bé hơn 2^2016

Đặt S = 1+10+102+103+...+102014

10S = 10 + 10² + 10³ +.......+ 10²⁰¹⁵

Ta có : 10S - S = ( 10 + 10² + 10³ +.......+10²⁰¹⁵) - ( 1 +10 + 10² + 10³ +........+10²⁰¹⁴)

9S = 10²⁰¹⁵ - 1

S = (10²⁰¹⁵ -1 ) :9

Ta thấy: ( 10²⁰¹⁵ - 1 ) : 9 < 10²⁰¹⁵

^{\(\Rightarrow\)} (1+10+102+103+...+102014) < 102015

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

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

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

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