tính S=(1-1/(1+2)).(1-1/(1+2+3)).(1-1/(1+2+3+4))......(1-1/(1+2+3+.........+2014))
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2014+\frac{2014}{1+2}+\frac{2014}{1+2+3}+...+\frac{2014}{1+2+3+...+10000}\)
\(S=\frac{2014}{\frac{1.2}{2}}+\frac{2014}{\frac{2.3}{2}}+\frac{2014}{\frac{3.4}{2}}+...+\frac{2014}{\frac{10000.10001}{2}}\)
\(S=\frac{4028}{1.2}+\frac{4028}{2.3}+\frac{4028}{3.4}+...+\frac{4028}{10000.10001}\)
\(S=4028\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10000.10001}\right)\)
\(S=4028\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10001-10000}{10000.10001}\right)\)
\(S=4028\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10000}-\frac{1}{10001}\right)\)
\(S=4028\left(1-\frac{1}{10001}\right)=\frac{40280000}{10001}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)
\(P=\frac{\left(1+2\right).2:2-1}{\left(1+2\right).2:2}.\frac{\left(1+3\right).3:2-1}{\left(1+3\right).3:2}.\frac{\left(1+4\right).4:2-1}{\left(1+4\right).4:2}...\frac{\left(1+2014\right).2014:2-1}{\left(1+2014\right).2014:2}\)
\(P=\frac{2}{2.3:2}.\frac{5}{3.4:2}.\frac{9}{4.5:2}...\frac{2029104}{2014.2015:2}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2013.2016}{2014.2015}\)
\(P=\frac{1.2.3...2013}{2.3.4...2014}.\frac{4.5.6...2016}{3.4.5...2015}\)
\(P=\frac{1}{2014}.\frac{2016}{3}=\frac{1}{2014}.672=\frac{336}{1007}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
cho a/b là phân số tối giản
chứng minh rằng a-2b/b cũng là phân số tối giản
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2014.2015}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(S=1-\frac{1}{2015}=\frac{2014}{2015}\)
Vậy:...
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2014}+\frac{1}{2015}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)=\frac{1}{1008}+\frac{1}{1009}+....+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Leftrightarrow S-P=0\) \(\Rightarrow\left(S-P\right)^{2016}=0\)
=[1*(1+2)/(1+2)-1/(1+2)]*...*[1*(1+2+3+...+2014)/(1+2+3+...+2014)-1/(1+2+3+...+2014]
còn lại là tịt
Câu hỏi của Phan Nguyễn Hà Linh - Toán lớp 6 - Học toán với OnlineMath