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

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

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

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)

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

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

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

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

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)

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

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

\(\Rightarrow A=\frac{2016}{2017}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)

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

\(\Rightarrow A=\frac{2016}{2017}\)

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= 1.2+2.3+3.4+...+2015.2016

3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3

    =1.2.3+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)

    =1.2.3-1.2.3+2.3.4-2.3.4+3.4.5+...-2014.2015.2016+2015.2016.2017

    =2015.2016.2017

A=2015.2016.2017:3=2731179360

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=\(\frac{2016}{2017}\)

mình quên ghi dấu "=" xin lỗi nhé

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

kếtq ủa dài lắm bạn nhan

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 :))