A  = 1.2.3 + 2.3.4 + ....+ 48.49.50

=> 4A = 1.2.3.4 + 2.3.4.(5-1) + ...+ 48.49.50.(51-17)

= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .....+ 48.49.50.51 - 47.48.49.50

= 48.49.50.51

=> A =  48.49.50.51:4 = 12.49.50.51

bài b) làm tương tự nha

3F= 1.2.(3-0)+ 2.3.(4-1)+...+ n.(n+1).[(n+2)-(n-1)] 
=[1.2.3+ 2.3.4+...+ (n-1)n(n+1)+ n(n+1)(n+2)]- [0.1.2+ 1.2.3+...+(n-1)n(n+1)] 
=n(n+1)(n+2) 
=>F 

H=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)

=> 4H=1.2.3(4-0)+2.3.4(5-1)+...+n(n+1)(n+2)((n+3)-(n-1))

=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+n(n+1)(n+2)(n+3)-(n-1).n(n+1)(n+2)

=n(n+1)(n+2)(n+3)

Nhân biểu thức S với số 5, ta có:

5.S = 1.2.3.4.5 + 2.3.4.5.5 + 3.4.5.6.5 + ... + 97.98.99.100.5

Biểu diễn số 5 ở mỗi số hạng vế phải bằng phép trừ thích hợp: 5 = 5 - 0 = 6 - 1 = 7 - 2 = ... = 101 - 96, ta có

5.S = 1.2.3.4.(5 - 0) + 2.3.4.5.(6 - 1) + 3.4.5.6.(7 - 2) + ...+ 97.98.99.100.(101 - 96)

     = (1.2.3.4.5 - 1.2.3.4.0) + (2.3.4.5.6 - 2.3.4.5.1) + (3.4.5.6.7 - 3.4.5.6.2) + ... + (97.98.99.100.101 - 97.98.99.100.96)

     = 1.2.3.4.5 - 0.1.2.3.4 + 2.3.4.5.6 - 1.2.3.4.5 + 3.4.5.6.7 - 2.3.4.5.6 + ... + 97.98.99.100.101 - 96.97.98.99.100

     = 97.98.99.100.101 - 0.1.2.3.4

     = 97.98.99.100.101

Suy ra  

  S = 97.98.99.100.101/5 = 97.98.99.20.101. Đến đây thì bạn dùng máy tính bấm ra S=1901009880

549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254

549 ,1326 ở đâu zậy bạn  !!! :/

Đặt A=1.2.3+2.3.4+.....+5.6.7

4A=1.2.3.4+2.3.4.(5-1)+.........+5.6.7.(8-4)

4A=1.2.3.4+2.3.4.5-1.2.3.4+........+5.6.7.8-4.5.6.7

4A=5.6.7.8

A=5.6.7.8:4

A=420

A=1(2+1)+2(3+1)+3(4+1)+...+99(100 +1 )

A=1.2+1+2.3+2+3.4+3...99.100+99

A=(1.2+2.3+3.4+...99.100)+(1+2+3+4...99)

giải:

Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101

4A=98.99.100.101

=>A=98.99.100.101/4

24497550

Ta có \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

Áp dụng:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{10\cdot11\cdot12}\\ =\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{10\cdot11}-\dfrac{1}{11\cdot12}\\ =\dfrac{1}{2}-\dfrac{1}{11\cdot12}=\dfrac{1}{2}-\dfrac{1}{132}=\dfrac{65}{132}\)

Ta có \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

Áp dụng

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{10\cdot11\cdot12}\\ =\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{10\cdot11\cdot12}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+..+\dfrac{1}{10\cdot11}-\dfrac{1}{11\cdot12}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{11\cdot12}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)=\dfrac{1}{2}\cdot\dfrac{65}{132}=\dfrac{65}{264}\)

Ta có: \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

Đặt \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)

   \(\Leftrightarrow2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\)

             \(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\)

             \(=\dfrac{1}{2}-\dfrac{1}{11.12}=\dfrac{65}{132}\)

  \(\Rightarrow A=\dfrac{65}{132}:2=\dfrac{65}{264}\)

A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)

3A-A= \(1-\frac{1}{3^{2008}}\)

B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}+\frac{1}{3^n}\)

3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-2}}+\frac{1}{3^{n-1}}\)

3B - B = \(1-\frac{1}{3^n}\)

Ta có :

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Leftrightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)

\(\Leftrightarrow\)\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(\Leftrightarrow\)\(2A=1-\frac{1}{3^{2008}}\)

\(\Leftrightarrow\)\(2A=\frac{3^{2008}-1}{3^{2008}}\)

\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{3^{2008}}:2\)

\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{2.3^{2008}}\)

Vậy \(A=\frac{3^{2008}-1}{2.3^{2008}}\)