1.2+2.3+3.4+....+2015.2016+2016.2017
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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
Đối với dạng này ta dùng công thức \(a\cdot\left(a+1\right)=\dfrac{1}{3}\left[a\cdot\left(a+1\right)\cdot\left(a+2\right)-\left(a-1\right)\cdot a\cdot\left(a+1\right)\right]\)
Ta có:
\(1\cdot2=\dfrac{1}{3}\left(1\cdot2\cdot3-0\cdot1\cdot2\right)\)
\(2\cdot3=\dfrac{1}{3}\left(2\cdot3\cdot4-1\cdot2\cdot3\right)\)
$\cdots$
\(2016\cdot2017=\dfrac{1}{3}\left(2016\cdot2017\cdot2018-2015\cdot2016\cdot2017\right)\)
Cộng lại ta có: \(1\cdot 2 +2\cdot 3 +3 \cdot 4 +\cdots +2016\cdot 2017=\dfrac{1}{3} (2016\cdot 2017 \cdot 2018-0\cdot 1 \cdot 2)=\dfrac{1}{3}\cdot 2016\cdot 2017 \cdot 2018 \)
Thay vào $A$ thu được $A=672.$
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+............+2015.2016.(2017-2014)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.............+2015.2016.2017-2014.2015.2016
3A=2015.2016.2017
A=2015.2016.2017:3
A=2015.672.2017
A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2017}\)
A=\(\frac{2016}{2017}\)
Đặt A = 1.2 + 2.3 + 3.4 + ... +2015.2016
3A = 1.2.3 + 2.3.(4-1) + ... + 2015.2016.(2017-2014)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 2015.2016.2017 - 2014.2015.2016
3A = 2014.2015.2016
A = 2727117120
4A= 4.( 1.2.3+2.3.4+4.5.6+5.6.7+...+2014.2015.2016
4A= 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .4.5.6.4 + ....+ 2014.2015.2016.4
4A= 1.2.3.4 + 2.3.4.(5 - 1)+ 3.4.5.( 6-2)+.....+ 2014.2015.2016.(2017-2013)
4A= 1.2.3.4+ 2.3.4.5-1.2.3.4.+ 3.4.5.6-2.3.4.5+ .....+ 2014.2015.2016.2017-2013.2014.2015.2016
4A = 2013.2014.2015.2016
A = 4117265071920
A=1.2+2.3+3.4+4.5+5.6+...+2016.2017
=> 3A = 1.2.3+2.3.3+3.4.3+4.5.3+5.6.3+.......+2016.2017.3
=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + 4.5.(6-3) + .......+ 2016.2017.(2018-2015)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +..........+ 2016.2017.2018 - 2015.2016.2017
=> 3A = 2016.2017.2018
=> A = 2016.2017.2018 : 3
Ta thấy:Các số trong dãy số trên cách nhau 1,1 đơn vị.
Số các số hạng là:
( 2016,2017 - 1,2 ) : 1,1 + 1 = 1832,819727 ( số )
Tổng là:
( 2016,2017 + 1,2 ) x 1832,819727 : 2 = 1848766,817
Đ/S: số trên dài wóa :))
ai giúp mình với
11.2+12.3+13.4+14.5+...+12015.2016+12016.2017
=1−12+12−13+13−14+14−15+...+12015−12016+12016−12017
=1−12017=20162017