So Sánh
\(A=\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2014}}+\frac{1}{5^{2015}}\)với \(\frac{1}{4}\)
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\(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2015}}\)
=>\(5A=1+\frac{1}{5}+...+\frac{1}{5^{2014}}\)
=>\(5A-A=\left(1+\frac{1}{5}+...+\frac{1}{5^{2014}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2015}}\right)\)
=>\(4A=1-\frac{1}{5^{2015}}\)
=>\(A=\frac{1-\frac{1}{5^{2015}}}{4}\)
Dễ thấy \(1-\frac{1}{5^{2015}}< 1\Rightarrow\frac{1-\frac{1}{5^{2015}}}{4}< \frac{1}{4}\Rightarrow A< \frac{1}{4}\)
\(S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}\)
\(5S=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}\)
\(\Rightarrow5S-S=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}\)
\(S=\frac{1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}}{4}\)
Xét \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
\(5A-A=1-\frac{1}{5^{2013}}\Leftrightarrow A=\frac{1-\frac{1}{5^{2013}}}{4}=\frac{1}{4}-\frac{1}{4.5^{2013}}\)
\(\Rightarrow S=\frac{1+\frac{1}{4}-\left(\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}\right)}{4}=\frac{5}{16}-\frac{\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}}{4}< \frac{1}{3}\)
A = 1/2.3/4.....2015/2016
= 1.3.5.....2015/2.4.6......2016
= 1.3.5.....2015/(1.2).(2.2).....(2.1008)
= 1.3.5.....2015/2^1008 . 1.2....1008
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)
\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)
\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)
ở tử số ta làm thế này
\(TS=\left(1+\frac{1}{2014}\right)+\left(1+\frac{1}{2013}\right)+\left(1+\frac{1}{2012}\right)+...+\left(1+\frac{2013}{2}\right)\)
\(TS=2015\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+...+\frac{1}{2}\right)\)
\(\frac{TS}{MS}=2015\)
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)