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\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2015.2016}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2015}-\frac{1}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)
3C=2014.2015.2016
C=2014.2015.2016:3
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Đặt A = 1 + 2 + 3 + 4 + ....... + 2014
Số các số hạng của A là:
2014 - 1 + 1 = 2014 (số)
A = 2014.(2014 + 1):2 = 2029105
Đặt B = 1.2 + 2.3 + 3. 4 + .....+ 2014.2015
3B = 1.2.3 + 2.3.3 + .3.4.3 + ......... + 2014.2015
3B = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + .......... + 2014.2015.(2016 - 2013)
3B = 1.2.3 + 2.3.4 - 1.2.3 + ........+2014.2015.2016 - 2013.2014.2015
3B = 2014 . 2015 .2016 = 8181351360
B = 8181351360 : 3
B = 2727117120
Vậy D = A + B = 2029105 + 2727117120 = 2729146225
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Đk: n khác 0, n khác -1
\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}\)
Vì \(0< \dfrac{1}{n+1}< 1\) (n khác 0, n khác -1) nên \(0< 1-\dfrac{1}{n+1}< 1\)
hay 0<S<1
Vậy S không là stn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\) (đpcm)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(\Rightarrow\) Quy đồng phân số và 1 là : \(\frac{49}{50}\) và \(1\)
Giữ nguyên phân số \(\frac{49}{50}\)
Ta có : \(\frac{1}{1}=\frac{1.50}{1.50}=\frac{50}{50}\)
\(\Rightarrow\frac{49}{50}< \frac{50}{50}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\); \(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(S=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); \(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)
=> \(S< \frac{3}{4}\)
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\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}\)
\(=\frac{2014}{2015}\)
A=1/1-1/2+1/2-1/3+1/3-...........+1/2014-1/2015
A=1/1-1/2015
A=2014/2015
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A = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100
A= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - ...... - 1/100
A = 1/1 - 1/100
A= 100/100 - 1/100
A= 99/100
A = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - ....... - 1/100
A= 1/1 - 1/100
A = 100 / 100 - 1/100
A= 99/100
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Phần chứng tỏ quy đồng lên rồi tính là ra
Còn phần tính S thì áp dụng tính chất vừa chứng tỏ để tách ra
Kết quả là 49/50
\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\)
\(S=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{2015-2014}{2014.2015}\)
\(S=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{2015}{2014.2015}-\frac{2014}{2014.2015}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(S=1-\frac{1}{2015}=\frac{2014}{2015}\)
\(\Rightarrow S