=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2011.2012.3

=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 2011.2012.( 2013 - 2010 )

=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + .... + 2011.2012.2013 - 2010.2011.2012

=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 2010.2011.2012 - 2010.2011.2012 ) + 2011.2012.2013

=> 3S = 2011.2012.2013

=> S = ( 2011.2012.2013 ) : 3

3S=1.2.3+2.3.(4-1)+...............+2011.2012.(2013-2010)

3S=1.2.3+2.3.4-1.2.3+...............+2011.2012.2013-2010.2011.2012

3S=2011.2012.2013

S=2011.2012.2013:3

S=2714954572

S=2714954572

       A = 1.2+2.3+3.4+.....+2011.2012

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ....... + 2011.2012.3

=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ........ + 2011.1012.(2013-2010)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ........... + 2011.2012.2013 - 2010.2011.2012

=> 3A = (1.2.3 + 2.3.4 + 3.4.5 + ........ + 2011.2012.2013) - (1.2.3 + 2.3.4 + 3.4.5 + ........ + 2010.2011.2012)

=> 3A = 2011.2012.2013

=> A = \(\frac{2011\cdot2012\cdot2013}{3}\)

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2011 - 1/2012

A = 1 - 1/2012

A = 2011/2012

B = 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 +...+ 1/2010 - 1/2012

B = 1/2 - 1/2012

B = 1005/2012

a) \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)

\(A=1-\dfrac{1}{2012}\)

\(A=\dfrac{2011}{2012}\)

b) \(B=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2010\cdot2012}\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2010\cdot2012}\right)\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2010}-\dfrac{1}{2012}\right)\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2012}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{1005}{2012}\)

\(B=\dfrac{1005}{4024}\)

Đặt S=1.2+2.3+.........+2011.2012

3S=1.2.3+2.3.(4-1)+...........+2011.2012.(2013-2010)

3S=1.2.3+2.3.4-1.2.3+...........+2011.2012.2013-2010.2011.2012

3S=2011.2012.2013

S=2011.2012.2013:3

S=2714954572

Đặt A = 1.2 + 2.3 + 3.4 + ... + 2011.2012

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2011.2012.3

=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2011.2012.(2013 - 2010)

=> 3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2011.2012.2013 - 2010.2011.2012

=> 3A = 2011.2012.2013

=> A = \(\frac{2011.2012.2013}{3}=2714954572\).

ket qua = 2714954572 ban nha

  A= 1.2 + 2.3 + 3.4 +...+ 2011.2012

=>3A= 1.2.3 + 2.3.3 + 3.4.3 +...+ 2011.2012.3

    3A= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ 2011.2012.(2013 - 2010)

    3A= (1.2.3 + 2.3.4 + 3.4.5 +...+ 2011.2012.2013) - (0.1.2 + 1.2.3 + 2.3.4 +...+ 2010.2011.2012)

    3A= (2011.2012.2013) - (0.1.2)

    3A= 8144863716 - 0

\(A=\frac{8144863716}{3}=2714954572\)

Vậy A= 2714954572

\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)

\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)

\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)

\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\)

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)

\(S=1-\frac{1}{2012}\)

\(S=\frac{2011}{2012}\)

Chúc bạn học tốt nha !!!

=1-1/2+1/2-1/3+1/3-1/4+...+1/2011-1/2012

= 1-1/2012

= 2011/2012

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\)

\(\Rightarrow S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)

\(\Rightarrow S=1-\frac{1}{2012}=\frac{2011}{2012}\)

1.50+2.49+3.48+...+49.2+50.1=

= (1.50+2.50+3.50+...+50.1)-(1.2+2.3+3.4+...+49.50)

= (2500+50).50:2-41650

= 63750-41650=22100

2, 

A = 1.2 + 2.3 + 3.4 + ... + 2011.2012

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2011.2012.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2011.2012.(2013 - 2010)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2011.2012.2013 - 2010.2011.2012

3A = 2011.2012.2013

A = 2011.2012.2013 : 3 

A = 2714954572

1)A=22100

2)B=60236

Cách 1:

Ta thấy mỗi số hạng của tổng trên là tích của hai số tự nhên liên tiếp, khi đó: 

Gọi a₁ = 1.2 → 3a₁ = 1.2.3 → 3a₁ = 1.2.3 - 0.1.2
      a₂ = 2.3 → 3a₂ = 2.3.3 → 3a₂ = 2.3.4 - 1.2.3
      a₃ = 3.4 → 3a₃ = 3.3.4 → 3a₃ = 3.4.5 - 2.3.4
      …………………..
      a_n-1 = (n - 1)n → 3a_n-1 =3(n - 1)n → 3a_n-1 = (n - 1)n(n + 1) - (n - 2)(n - 1)n
      a_n = n(n + 1) → 3a_n = 3n(n + 1) → 3a_n = n(n + 1)(n + 2) - (n - 1)n(n + 1)

Cộng từng vế của các đẳng thức trên ta có:

3(a₁ + a₂ + … + a_n) = n(n + 1)(n + 2)

Cách 2: Ta có

3A = 1.2.3 + 2.3.3 + … + n(n + 1).3 = 1.2.(3 - 0) + 2.3.(3 - 1) + … + n(n + 1)[(n - 2) - (n - 1)] = 1.2.3 - 1.2.0 + 2.3.3 - 1.2.3 + … + n(n + 1)(n + 2) - (n - 1)n(n + 1) = n(n + 1)(n + 2) 

* Tổng quát hoá ta có:

k(k + 1)(k + 2) - (k - 1)k(k + 1) = 3k(k + 1). Trong đó k = 1; 2; 3; …

Ta dễ dàng chứng minh công thức trên như sau:

k(k + 1)(k + 2) - (k - 1)k(k + 1) = k(k + 1)[(k + 2) - (k - 1)] = 3k(k + 1)