cho S=1-1/2+1/3-1/4+...+1/2011-1/2012+1/2013 và P=1/1007+1/1008+...+1/2013
Tính (s-P)^2013
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TB
16 tháng 4 2017
S-P= (1 - 1/2 + 1/3 - 1/4 +...+ 1/2011 - 1/2012 + 1/2013) - ( 1/1007 + 1/1008 +...+ 1/2012 + 1/2013 )
S-P= (1- 1/2 + ... + 1/1005 - 1/1006) - 2.(1/1008 + 1/1010 + 1/1012 +...+ 1/2012)
S-P= 1+1/2+1/3+...+1/1006 - 2.( 1/2 + 1/4 + 1/6 +...+ 1/2012)
S-P= 1 + 1/2 + 1/3 +...+ 1/1006 - ( 1+ 1/2 + 1/3 +...+ 1/1006 )
S-P= 0
Suy ra (S-P)^2013 = 0
21 tháng 8 2019
link này nè bn!
https://olm.vn/hoi-dap/detail/103540952175.html
21 tháng 8 2019
S-P= (1 - 1/2 + 1/3 - 1/4 +...+ 1/2011 - 1/2012 + 1/2013) - ( 1/1007 + 1/1008 +...+ 1/2012 + 1/2013 )
S-P= (1- 1/2 + ... + 1/1005 - 1/1006) - 2.(1/1008 + 1/1010 + 1/1012 +...+ 1/2012)
S-P= 1+1/2+1/3+...+1/1006 - 2.( 1/2 + 1/4 + 1/6 +...+ 1/2012)
S-P= 1 + 1/2 + 1/3 +...+ 1/1006 - ( 1+ 1/2 + 1/3 +...+ 1/1006 )
S-P= 0
(S-P)^2013 = 0
27 tháng 4 2016
S = 1/3+1/5+1/7+...+1/2013-(1/2+1/4+1/6+...+1/2012)
S = 1/2+1/3+1/4+...+1/2012+1/2013 - 2(1/2+1/4+1/6+...+1/2012)
S = 1/2+1/3+1/4+...+1/2012+1/2013 - (1+1/2+1/3+...+1/1006)
S = 1/1007+1/1008+...+1/2013-1
=> S - P = 1/1007+1/1008+...+1/2013-1-(1/1007+1/1008+...+1/2013)
<=> S - P= -1 <=> (S-P)2013 = -1
H
0
10 tháng 4 2017
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1006}\)
\(S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}=P\)
=>(S-P)2013=02013=0
2 tháng 4 2019
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NT
2 tháng 4 2019
Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{1006}\)
\(=\frac{1}{1007}+\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2012}+\frac{1}{2013}\left(1\right)\)
Mà \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow S=P\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
Vậy...
7 tháng 4 2018
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{3}+......+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+........+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+......+\frac{1}{2013}\)
\(=P\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0\)
15 tháng 3 2020
Ta có :
\(S=\left(1+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}=P\)
\(\Rightarrow\left(s-p\right)^{2013}=0^{2013}=0\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}\right)\)
\(S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)
\(\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\).
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