cho S=1-1/2+1/3-1/4+......+1/2015-1/2016+1/2017 và P=1/1009+1/1010+.....+1/2016+1/2017
Tính (S-P)^2016
K
Khách
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Những câu hỏi liên quan
CP
0
TH
0
NN
0
HT
1
6 tháng 9 2017
\(M=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}\)
\(M=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)\(M=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)\(M=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2008}\right)\)
\(M=\dfrac{1}{2009}+\dfrac{1}{2010}+...+\dfrac{1}{2016}+\dfrac{1}{2017}=N\)
Vậy \(\left(M-N\right)^{2017}=0\)
30 tháng 7 2016
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
HP
14 tháng 5 2016
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
14 tháng 5 2016
Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!
LQ
1
5 tháng 5 2016
Xét số chia: 1-\(\frac{1}{2}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) +...+\(\frac{1}{2015}\) - \(\frac{1}{2016}\)
= (1+\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) +...+\(\frac{1}{2015}\) + \(\frac{1}{2016}\)) - 2.(\(\frac{1}{2}\) + \(\frac{1}{4}\) + ... + \(\frac{1}{2016}\))
= (1+\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) +...+\(\frac{1}{2015}\) + \(\frac{1}{2016}\)) - (1+\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) +...+\(\frac{1}{1007}\) + \(\frac{1}{1008}\))
=\(\frac{1}{1009}\) + \(\frac{1}{1010}\) + ... + \(\frac{1}{2015}\)+ \(\frac{1}{2016}\) => A=1
