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
Những câu hỏi liên quan
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
(1/1008 + 1/1009 + 1/1010 +...+ 1/2014) : [1/2 + 1/(3.4) +1/(5.6) +...+ 1/(2013.2014)]=???
CMR :1/1.2+1/3.4+...+1/2013.2014=1/1008+1/1009 +........+1/2013+1/2014
Chứng minh rằng : \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\)
cho A=1/1.2+1/3.4+1/5.6+....+1/2021.2022 và B=1011+1010/1012+1009/1013+1008/1014+...+2/2020+1/2021 Chứng minh rằng : B/A là số nguyên
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 1/1.2+1/3.4+1/5.6+...+1/2015.2016=1/1009+1/1010+...+1/2016
Tính B=1/1008*2014 + 1/1009*2013 + 1/1010*2012 + ...+ 1/2014*1008
Chứng minh: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\).