A= 1.2+2.3+3.4+...+2015.2016

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

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

3A=8193538080

A=8193538080:3

A=2731179360

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

=> 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 + .... + 2015.2016.2017 - 2014.2015.2016

=> 3A = 2015.2016.2017

=> A = \(\frac{2015.2016.2017}{3}\)

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

Đặt \(A=1.2+2.3+3.4+...+2015.2016\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2015.2016.\left(2017-2014\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2015.2016.2017-2014.2015.2016\)

\(\Rightarrow3A=2015.2016.2017\)

\(\Rightarrow A=2015.2016.2017:3\)

\(\Rightarrow A=2015.672.2017\)

Vậy \(A=2015.672.2017\)

1 . 2 + 2 . 3 + 3 . 4 + ... + 2015 . 2016

3M = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 2015 . 2016 . 3 

3M = 1 . 2 ( 3 - 0 ) + 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 2015 . 2016 ( 2017 - 2014 )

3M = ( 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4. 5 + ... + 2015 . 2016 . 2017 ) - ( 0 . 1 . 2 + 1 . 2 . 3 + 2 . 3 . 4 + ... + 2014 . 2015 . 2016 )

3M = 2015 . 2016 . 2017

M = \(\frac{2015.2016.2017}{3}\)

M = 2731179360

Gọi tổng 1.2+2.3+3.4+...+2015.2016 là M

\(\text{M = 1.2+2.3+3.4+...+2015.2016}\)

\(3M=1.2.3+2.3.3+3.4.3+...+2015.2016.3\)

\(3M=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+2015.2016.\left(2017-2014\right)\)

\(3M=\left(1.2.3+2.3.4+3.4.5+...+2015.2016.2017\right)-\left(0.1.2+1.2.3+2.3.4+...+2014.2015.2016\right)\)

\(M=2015.2016.2017\)

\(M=\frac{2015.2016.2017}{3}\)

\(M=672.2015.2017\)

\(M=2731179360\)

a) \(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{10200}\)

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)

\(2A=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+...+\left(\frac{1}{100}-\frac{1}{102}\right)\)

\(2A=\frac{1}{2}-\frac{1}{102}\)

\(2A=\frac{25}{51}\)

\(A=\frac{25}{51}:2\)

\(A=\frac{25}{102}\)

Vậy \(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{10200}=\frac{25}{102}\)

b) \(B=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}\)

\(B=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)

\(B=3.\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\right]\)

\(B=3.\left(\frac{1}{1}-\frac{1}{2016}\right)\)

\(B=3.\frac{2015}{2016}\)

\(B=\frac{2015}{672}\)

Vậy \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}=\frac{2015}{672}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

 Tính tổng :

a ,1+2+3+..........+2015 

SSH của tổng trên là :

   (2015-1):1+1=2015(SH)

Tổng trên là:

  (2015:2)x(2015+1)=2031120

b, 3+5+7+......+2015

SSH của tổng trên là :

     (2015-3):2+1=1007(SH)

Tổng trên là:

     (1007:2)x(2015+3)=1016063

LƯU ý: SSH=số số hạng nha

a 2029106

b508032

c1679780.53381924

tick đúng cho mk nha

                         a)1+2+3+...........+2015 = 2031120

                          b)3+5+7+...........+2015 = 1016063

                          c)1,2+2,3+3,4+...........2015,2016 =20294103,83

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

\(S=\dfrac{3}{1.2}+\dfrac{3}{2.3}+...+\dfrac{3}{2015.2016}\)

\(=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2015.2016}\right)\)

\(=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

\(=3\left(1-\dfrac{1}{2016}\right)\)

\(=3.\dfrac{2015}{2016}=\dfrac{6045}{2016}\)

Vậy \(S=\dfrac{6045}{2016}\)

\(S=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)

\(\Rightarrow S=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

\(\Rightarrow S=3\left(1-\dfrac{1}{2016}\right)=3.\dfrac{2015}{2016}=\dfrac{6045}{2016}\)

Vậy ...

\(S=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+....+\dfrac{3}{2015.2016}\)

\(=\dfrac{1}{1}.\left(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+....+\dfrac{3}{2015.2016}\right)\)

= \(3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....\dfrac{1}{2015.2016}\right)\)

= 3. \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\)

= 3.