\(A=\frac{1}{19}+\frac{2}{19^2}+...+\frac{2014}{19^{2014}}\)

\(\Rightarrow19A=1+\frac{2}{19}+\frac{3}{19^2}+...+\frac{2014}{19^{2013}}\)

\(\Rightarrow19A-A=\left(1+\frac{2}{19}+\frac{3}{19^2}+...+\frac{2014}{19^{2013}}\right)-\left(\frac{1}{19}+\frac{2}{19^2}+...+\frac{2014}{19^{2014}}\right)\)

\(\Rightarrow18A=1+\left(\frac{1}{19}+\frac{1}{19^2}+...+\frac{1}{19^{2013}}\right)-\frac{2014}{19^{2014}}\)

\(\Rightarrow18A=1+\frac{1-\frac{1}{19^{2013}}}{18}-\frac{2014}{19^{2014}}\)

\(\Rightarrow A=\frac{1+\frac{1-\frac{1}{19^{2013}}}{18}-\frac{2014}{19^{2014}}}{18}\)

Vậy...

Không cần giải cũng biết đáp án:

Nếu A là số dương thì A^2016>A^2015

Nếu A là số âm thì A^2016 là số dương , A^2015 là số âm nên chắc chắn A^2016>A^2015

k nha

quy dong ca A va B ta dc :

\(A=\frac{-109}{10^{2014}}\)

\(B=\frac{-199}{10^{2014}}\)

\(\Rightarrow A>B\)

dễ thôi

ta có :A=-9/10^2013+-19/10^2014=-9/10^2013+-9/10^2014+-10/10^2014

          B=-9/10^2014+-19/10^2013=-9/10^2014+-9/10^2013+-10/10^2013

nhìn nhé :cả A và B đều có các số hạng :-9/10^2013 và-9/10^2014

mà -10/10^2014<-10/10^2013

=>A<B

\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+....+\frac{2014}{4^{2014}}\)

\(4S=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\)

\(4S-S=\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\right)\)

\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)

\(12S=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\)

\(12S-3S=\left(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\right)\)

\(9S=4-\frac{2014}{4^{2013}}-\frac{1}{4^{2013}}+\frac{2014}{4^{2014}}\)

\(9S=4-\frac{4028}{4^{2014}}-\frac{4}{4^{2014}}+\frac{2014}{4^{2014}}\)

\(9S=4-\frac{2010}{4^{2014}}< 4\)

\(\Rightarrow9S< 4\)

\(\Rightarrow S< \frac{4}{9}< 1\)(đpcm)

Ta có :

\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)( 1 )

\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2014}{4^{2013}}\)( 2 )

Lấy ( 2 ) - ( 1 ) ta được :

\(3S=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)

gọi     \(B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)( 3 )

\(4B=4+1+\frac{1}{4}+...+\frac{1}{4^{2012}}\)  ( 4 )

Lấy ( 4 ) - ( 3 ) ta được :

\(3B=4-\frac{1}{4^{2013}}\)

\(\Rightarrow B=\frac{4-\frac{1}{4^{2013}}}{3}=\frac{4}{3}-\frac{1}{4^{2013}.3}\)

\(\Rightarrow3S=\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}\)

\(\Rightarrow S=\frac{\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}}{3}=\frac{4}{9}-\frac{1}{4^{2013}.9}-\frac{2014}{4^{2014}.3}< \frac{4}{9}< 1\)

vậy \(S< 1\)