a=1.2 + 2.3 +3.4+ ...+ 2010.2011\

3a=1.2.3+2.3.3+3.4.3+......+2010.2011.3

3a=1.2.3+2.3.(4-1)+3.4.(5-1)+............+2010.2011.(2012-2009)

3a=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+2010.2011.2012-2009.2010.2011

3a=2010.2011.2012

a=2010.2011.2012:3

a=?

A=2010.2011.2012:3

Ta có \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}\)

\(\Rightarrow A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)

\(\Rightarrow A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2016}+\frac{1}{2017}\right)\)

\(\Rightarrow A=2\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=2\left(\frac{2016}{2017}\right)\)

\(\Rightarrow A=\frac{4032}{2017}\)

Ta có:\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2016\cdot2017}\)

\(=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+....+\frac{2}{2016}-\frac{2}{2017}\)

\(=\frac{2}{1}-\frac{2}{2017}=2-\frac{2}{2017}=\frac{4034}{2017}-\frac{2}{2017}=\frac{4032}{2017}\)

\(A=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{2016\cdot2017}\)

\(\frac{A}{2}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)

\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(\frac{A}{2}=1-\frac{1}{2017}=\frac{2016}{2017}\)

\(A=\frac{2016}{2017}\cdot2=\frac{4032}{2017}\)

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 $A=672.$

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

A=1.2.3+2.3.4+...2016.2017.2017-2.3.4+.....2015.2016.2017

A=1.2017=2017 :D làm sai nhá

Trần Đức Hùng lần đầu t soi bài m Hùng xinh gái ak :>

\(A=1.2+2.3+3.4+4.5+...+2016.2017\)

\(3A=1.2.3+2.3.\left(4-1\right)+...+2016.2017.\left(2018-2015\right)\)

\(3A=1.2.3+2.3.4-1.2.3+...+2016.2017.2018-2015.2016.2017\)

\(3A=2016.2017.2018\Rightarrow A=\frac{2016.2017.2018}{3}\)

p/s: lần sau lèm cẩn thận nha bn iu dấu, để mấy em lớp 6 bt nhục mặt vl :D

cc ok nhé :D tú vụm

\(A=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right):2\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right):2\)

\(=\left(1-\frac{1}{2017}\right):2\)\(< \)\(\frac{1}{2}\)   (Do 1 - 1/2017 < 1)