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\(\text{đặt}k=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(K=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+....+\frac{1}{2017}\Rightarrow A=1\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(S=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow\left(S-P\right)^{2016}=\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{1009}-...-\frac{1}{2015}\right)^{2016}=0^{2016}=0\)
Ta thấy:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)+\frac{1}{2015}\)
\(S=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\) nên:
\(S=P\)\(\Rightarrow S-P=0\)\(\Rightarrow\left(S-P\right)^{2016}=0\)
Đặt A là tên của biểu thức
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}-\left(1+\frac{1}{2}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}+\frac{1}{2017}\)
Do đó \(A=\frac{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2017}}{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2017}}=1\)
Ta có \(B=\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}\)
\(\Rightarrow B=1+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)\)
\(\Rightarrow B=\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}\)
\(\Rightarrow B=2017.\left(\frac{1}{2017}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)
\(\Rightarrow B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)= 2017
~ Chúc bạn học tốt
khó was chịu mk cũng lớp 7 mà chẳng thấy bài nào như vạy cả
bài này dễ mà? tớ lười quá chẳng muốn làm đâu