c: C=1*2+2*3+3*4+...+58*59+59*60

=>3*C=1*2*3+2*3*(4-1)+3*4*(5-2)+...+58*59(60-57)+59*60(61-58)

=>3*C=1*2*3+2*3*4-1*2*3+...+58*59*60-58*59*57+59*60*61-58*59*60

=>3*C=59*60*61

=>C=59*20*61=71980

\(1+\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+\frac{7}{3\cdot4}+...+\frac{7}{59\cdot60}\)

\(=1+7\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{59\cdot60}\right)\)

\(=1+7\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(=1+7\left(1-\frac{1}{60}\right)\)

\(=1+7\cdot\frac{59}{60}\)

Cảm ơn bạn nha Sooya.

\(A=1.2+2.3+3.4+4.5+...+59.60\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+59.60.\left(61-58\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+...+59.60.61-58.59.60\)

\(\Rightarrow3A=59.60.61\)

\(\Rightarrow A=\frac{59.60.61}{3}\)

b: Tổng của N là:

\(\dfrac{49\cdot48}{2}=49\cdot24=1176\)

chào nick thứ 2 đây

a) \(3M=1.2.3+2.3.3+...+48.49.3=1.2.3+2.3.\left(4-1\right)+...+48.49.\left(50-47\right)=1.2.3+2.3.4-1.2.3+...+48.49.50-47.48.49=48.49.50\Rightarrow M=\dfrac{48.49.50}{3}\Rightarrow M=39200\)

b) Tương tự câu a

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34

=> 3S = 32.33.34

=> S = \(\frac{32.33.34}{3}=11968\)

Tk:
Đặt P = 1.2+2.3+3.4+...+99.100

3P = 1.2.3+2.3.3+3.4.3+...+99.100+3

3P = 1.2 (3-0) +2.3(4-1)+3.4(5-2) +...+ 99.100( 101-98)

3P = ( 1.2.3 + 2.3.4 + 3.4.5 + 99.100.101 ) -( 0.1.2 + 1.2.3 + 2.3.4 + ....+ 98.99.100)

3P = 99.100.101 - 0.1.2

3P = 999900 - 0

3P = 999900

P = 999900 : 3

P = 333300

\(A=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3A=1.2.3+2.3.3+...+99.100.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4\left(5-2\right)+...+99.100\left(101-98\right)\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-....-98.99.100+99.100.101\)

\(=99.100.101\)

\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)

 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

Số các số hạng của A là :

( 99 - 1 ) : 1 + 1 = 99 ( số )

Tổng A là :

( 99 + 1 ) . 99 : 2 = 4950

Vậy tổng A = 4950 .

Bạn nhân 3 lên

Mk ko ghi lại đề bài mà làm tiếp luôn nhé!

3A=1.2.3+2.3.3+3.4.3+.......+98.99.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+.........+98.99.(100-97)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..............

3A=98.99.100

A=98.99.100/3

Bn k cho mk nhé!

B= 

a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)