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\(M=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}\)
\(M=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)\(M=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)\(M=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2008}\right)\)
\(M=\dfrac{1}{2009}+\dfrac{1}{2010}+...+\dfrac{1}{2016}+\dfrac{1}{2017}=N\)
Vậy \(\left(M-N\right)^{2017}=0\)
Theo đầu bài ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow S=P\)
Vậy ( S - P )2016 = 02016 = 0
\(N=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(N=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+...+\frac{1}{2016}\right)\)
\(N=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1008}\right)\)
\(N=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}=K\)