CMR 1/1.2+1/3.4+1/5.6+...+1/2015.2016=1/1009+1/1010+...+1/2016
Những câu hỏi liên quan
chứng minh 1/1.2+1/3.4+1/5.6+...+1/2015.2016=1/1009+1/1010+...+1/2016
Cho A=1/1.2+1/3.4+1/5.6+...+1/2015.2016 và B=1/1008+2/1009+1/1010+...+1/2016. Tính B-A
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2015}+\frac{1}{2016}\)
\(\Rightarrow B-A=\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\right)\)
\(\Rightarrow B-A=\frac{1}{1008}\)
Chứng minh: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\).
CM:
1/1.2 + 1/3.4 + 1/5.6 + ...+ 1/2015.2016 = 1/1009 + 1/1010 + ... + 1/2016
cho A=1/1.2+1/3.4+1/5.6+.....+1/2015.2016 và B= 1/1008+1/1009+....+1/2016
tính A-B
CMR: 1/1.2+1/3.4+1/5.6+1/7.8+...+1/2013.2014=1/1008+1/1009+1/1010+...+1/2013+1/2014
CMR:1/2+1/3.4+1/5.6+...+1/2015.2106=1/1009+1/1010+...+1/2016
\(CMR:\dfrac{1}{2}+\dfrac{1}{3}.4+\dfrac{1}{5}.6+...+\dfrac{1}{2015}.2016=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2016}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2016}\left(đpcm\right)\).
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Chứng minh :
\(\frac{1}{2}+\frac{1}{3.4}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Đặt tổng là S
\(\Rightarrow S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\)
\(\Rightarrow S=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(\Rightarrow S=\left(1+\frac{1}{2}+....+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(\Rightarrow S=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\) (đpcm)
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Chứng minh rằng
1/1.2 + 1/3.4 + 1/5.6 + 1/7.8 + ... + 1/2013.2014 = 1/1008 + 1/1009 + 1/1010 +...+ 1/2013+ 1/2014
$2015=5.13.31$2015=5.13.31
Ta có: $1.2.....1007=1.2...5....13.....31...1007\text{ chia hết cho }5.13.31=2015$1.2.....1007=1.2...5....13.....31...1007 chia hết cho 5.13.31=2015
$1008.1009.....2004=1008....\left(1010\right)....\left(1014\right)...\left(1023\right)....2004$1008.1009.....2004=1008....(1010)....(1014)...(1023)....2004
$=1008....\left(5.202\right)....\left(13.78\right)....\left(31.33\right)...2004\text{ chia hết cho }5.13.33=2015$=1008....(5.202)....(13.78)....(31.33)...2004 chia hết cho 5.13.33=2015
Do đó tổng 2 số trên chia hết cho 2015.
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