3/1.2.3+5/2.3.4+...+2017/1008+1009+1010<5/4
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VA
0
12 tháng 4 2016
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
NT
0
29 tháng 4 2016
Theo mình thì đề bài đầy đủ là như thế này :
So sánh \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2015\cdot2016}\)với \(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\).
Giải :
Ta có : \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2015\cdot2016}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\cdot2\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}< \frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Chúc bạn học tốt!
S
8 tháng 6 2017
Sửa đề: Cho \(V=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)và \(Y=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\). Tính \(\frac{V}{Y}\)
\(V=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
=> \(\frac{V}{Y}=\frac{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}=1\)
8 tháng 6 2017
V = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
V = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
V = \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
V = \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
V = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
V = \(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Vậy V : Y = \(\frac{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}{\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2016}}\)
( Mình nghĩ Y = 1/1009 + 1/1010 + ... + 1/2016 / Nếu Y như mình nói thì V : Y = 1 )
TT
0