\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\\ =1-\dfrac{1}{2014}\\ =\dfrac{2013}{2014}\)

\(=\frac{2013}{2014}\)

\(\frac{2013}{2014}\)

2013/2014

\(\frac{2013}{2014}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2013.2014}.\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2013}-\frac{1}{2014}\)

\(A=1-\frac{1}{2014}\)

\(A=\frac{2013}{2014}\)

câu a) (a^2+2a+a+2)(a+3)-(a^2+a)(a+2)= (3a+3)(a+2)

suy ra: a^3+3x^2+2a^2+6a+a^2+3a+2a+6-a^3-2x^2-a^2-2a= 3a^2+6a+3a+6

3a^2+9a+6=3a^2+9a+6

câu b) 

^ là gì vậy bạn

 A = 1.2 + 2.3 + 3.4 + ... + 2013.2014 
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2013.2014.3 
Mà : 
1.2.3 = 1.2.3 
2.3.3 = 2.3.4 - 2.3.1 
3.4.3 = 3.4.5 - 3.4.2 

2012.2013.3 = 2012.2013.2014 - 2012.2013.2011 
2013.2014.3 = 2013.2014.2015 - 2013.2014.2012 
Cộng tất cả, vế theo vế ---> 3S = 2013.2014.2015 
---> A = 2013.2014.2015 / 3 = 2723058910.

của bạn đây

Lời giải:

$3A=1.2.3+2.3(4-1)+3.4(5-2)+...+2013.2014(2015-2012)$

$=(1.2.3+2.3.4+3.4.5+...+2013.2014.2015)-(1.2.3+2.3.4+...+2012.2013.2014)$

$=2013.2014.2015$

$\Rightarrow A=\frac{2013.2014.2015}{3}$

 A = 1.2 + 2.3 + 3.4 + ... + 2013.2014 
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2013.2014.3 
Mà : 
1.2.3 = 1.2.3 
2.3.3 = 2.3.4 - 2.3.1 
3.4.3 = 3.4.5 - 3.4.2 

2012.2013.3 = 2012.2013.2014 - 2012.2013.2011 
2013.2014.3 = 2013.2014.2015 - 2013.2014.2012 
Cộng tất cả, vế theo vế ---> 3S = 2013.2014.2015 
---> A = 2013.2014.2015 / 3 = 2723058910

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

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2013.2014.3

=> 3A = 1.2.( 3 - 0 ) + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ... + 2013.2014.( 2015 - 2012 )

=> 3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 2013.2014.2015 - 2012.2013.2014

=> 3A = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 2012.2013.2014 - 2012.2013.2014 ) + 2013.2014.2015

=> 3A = 2013.2014.2015

=> A = ( 2013.2014.2015 ) : 3

a= 1.2 + 2.3 + 3.4 + 4.5 + ... + 2013.2014
3a = 1.2.3 +2.3.3 +3.4.3 +4.5.3 +... + 2013.2014.3
3a = 1.2.(3-0)+2.3(4-1)+3.4.(5-2)+...+2013.2014.(2015-2012)
3a = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2013.2014.2015-2012.2013.2014
3a=2013.2014.2015
a = 2013.2014.20153
a = 2723058910

\(\)Đáp án: \(\dfrac{2013.2014.2015}{3}\)

Tổng quát: \(S_n=1.2+2.3+...\left(n-1\right).n\)

Ta sẽ chứng minh \(S_n=\dfrac{\left(n-1\right)n\left(n+1\right)}{3}\) với mọi n nguyên, \(n\ge2\) bằng quy nạp.

- Với \(n=2:S_2=1.2=2=\dfrac{1.2.3}{3}\)

- Giả sử khẳng định đúng đến \(n=k:S_k=\dfrac{\left(k-1\right)k\left(k+1\right)}{3}\)

- Với \(n=k+1:\)

 \(S_{k+1}=1.2+2.3+...+\left(k-1\right).k+k.\left(k+1\right)\\ =S_k+k.\left(k+1\right)\\ =\dfrac{\left(k-1\right).k.\left(k+1\right)}{3}+k.\left(k+1\right)\\ =\dfrac{\left(k-1\right).k.\left(k+1\right)+3.k.\left(k+1\right)}{3}\\ =\dfrac{k.\left(k+1\right).\left(k+2\right)}{3}\left(\text{dpcm}\right)\)

Vậy \(D=S_{2014}=\dfrac{2013.2014.2015}{3}\)