1/1x2+1/2x3+1/3x4+...+1/2012+2013
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![](https://rs.olm.vn/images/avt/0.png?1311)
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A = 1×2 + 2×3 + 3×4 + ... + 98×99
3A = 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 98×99×(100-97)
3A = 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 98×99×100 - 97×98×99
3A = 98×99×100
A = 98×33×100
A = 323400
2) Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
Ta có:
102012 + 1/102013 + 1 < 102012 + 1 + 9/102013 + 1 + 9
< 102012 + 10/102013 + 10
< 10.(102011 + 1)/10.(102012 + 1)
< 102011 + 1/102012 + 1
Vào lúc: 2016-07-17 13:22:30 Xem câu hỏi
1) A = 1×2 + 2×3 + 3×4 + ... + 98×99
3A = 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 98×99×(100-97)
3A = 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 98×99×100 - 97×98×99
3A = 98×99×100
A = 98×33×100
A = 323400
2) Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
Ta có:
102012 + 1/102013 + 1 < 102012 + 1 + 9/102013 + 1 + 9
< 102012 + 10/102013 + 10
< 10.(102011 + 1)/10.(102012 + 1)
< 102011 + 1/102012 + 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}=\frac{2013}{2014}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2013}{2014}\)
\(\Rightarrow1-\frac{1}{n+1}=\frac{2013}{2014}\)
\(\Rightarrow\frac{1}{n+1}=1-\frac{2013}{2014}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{2014}\)
\(\Rightarrow n+1=2014\)
\(\Rightarrow n=2014-1\)
\(\Rightarrow n=2013\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2013}{2014}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\Rightarrow A=1-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2013}{2014}\)
\(\Rightarrow\)\(\frac{1}{x+1}=\frac{1}{2014}\)
\(\Rightarrow x+1=2014\)
\(\Rightarrow x=2014-1\)
\(\Rightarrow x=2013\)
Vậy x=2013
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)
\(1-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\frac{1}{x+1}=1-\frac{2013}{2014}\)
\(\frac{1}{x+1}=\frac{1}{2014}\)
Vì \(x+1\)là mẫu số nên:
\(x+1=2014\)
\(x=2014-1=2013\)
Vậy ....
P/s: Dấu . là nhân nha!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
= 1 -1/2 + 1/2 -1/3 + 1/3 - 1/4 + .....+ 1/2012 - 1/2013
= 1-1/2013
=2012/2013
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