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Ta có: A = 1 - 1/2 + 1/3 - 1/4 + ... +1/2013 - 1/2014
A = 1 + 1/2 + 1/3 + 1/4 +... + 1/2013 + 1/2014 - 2.(1/2 + 1/4 + ... + 1/2014)
A = 1 + 1/2 + 1/3 + 1/4 +... + 1/2013 + 1/2014 - (1 + 1/2 + 1/3 + ... + 1/1007)
A = 1/1008 + 1/1009 + ... + 1/2014
bạn viết lại B được ko
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(A=1-\frac{1}{2014}\)
\(A=\frac{2013}{2014}\)
bài B thì đề khó hiểu quá
bn ghi lại đề rồi mình giải
$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.
Sửa đề: \(\frac{\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}}{\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}}\)
Ta có: \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}\)
\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2013}-\frac{1}{2014}\)
\(=1+\frac12+\frac13+\cdots+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac12+\frac14+\cdots+\frac{1}{2014}\right)\)
\(=1+\frac12+\frac13+\cdots+\frac{1}{2013}+\frac{1}{2014}-1-\frac12-\cdots-\frac{1}{1007}\)
\(=\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2014}\)
Ta có: \(\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}\)
\(=\frac{2}{1008\cdot2014}+\frac{2}{1009\cdot2013}+\cdots+\frac{2}{1510\cdot1512}+\frac{1}{1511\cdot1511}\)
\(=2\left(\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{1510\cdot1512}\right)+\frac{1}{1511\cdot1511}\)
\(=\frac{2}{3022}\left(\frac{3022}{1008\cdot2014}+\frac{3022}{1009\cdot2013}+\cdots+\frac{3022}{1510\cdot1512}\right)+\frac{1}{1511\cdot1511}\)
\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+\cdots+\frac{1}{1510}+\frac{1}{1512}\right)+\frac{1}{1511}\cdot\frac{1}{1511}\)
\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2013}+\frac{1}{2014}\right)\)
Ta có: \(\frac{\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}}{\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}}\)
\(=\frac{\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2014}}{\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2013}+\frac{1}{2014}\right)}\)
\(=1:\frac{1}{1511}=1511\)