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\(M=5^{2016}-5^{2015}-5^{2014}-...-5-1\)
=>\(5M=5\left(5^{2016}-5^{2015}-5^{2014}-...-5-1\right)\)
=>\(5M=5^{2017}-5^{2016}-5^{2015}-...-5^2-5\)
=>\(5M-M=\left(5^{2017}-5^{2016}-5^{2015}-...-5^2-5\right)-\left(5^{2016}-5^{2015}-5^{2014}-...-5-1\right)\)
=>\(4M=5^{2017}-2.5^{2016}+1\)
=>\(M=\frac{5^{2017}-2.5^{2016}+1}{4}\)
M = 52016 - (52015 + 52014 + ... + 5 + 1) = 52016 - (52016 - 1) = 52016 - 52016 + 1 = 0 + 1 = 1
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)
\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)
\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)
\(5^{2016}-5^{2015}-....-5-1\)
\(=5^{2016}-\left(5^{2015}+5^{2014}+....+5+1\right)\)
\(=5^{2016}-\left(5^{2016}-1\right)\)
\(=5^{2016}-5^{2016}+1\) \(=0+1=1\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>5M=5(52016−52015−52014−...−5−1)
=>5M=52017−52016−52015−...−52−5
=>5M−M=(52017−52016−52015−...−52−5)−(52016−52015−52014−...−5−1)
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