có 2 cách bạn ạ 

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)

học tốt 

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)

tham khảo trên mạng có cả !!

     1.2 + 2.3 + 3.4+...+n.( n+1)=A

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

 =>3.A=1.2.3+2.3.4+3.4.5+...+ (n-1) . n. (n+1)+ n. (n+1). (n+2) - 0.1.2 -1.2.3 -2.3.4 -3.4.5 -...(n-1)n(n+1)

  =>3A=n.(n+1).(n+2)

  => A=n.(n+1).(n+2)\3 

Đặt A=1.2 + 2.3 + 3.4+...+n.( n+1)

=>3A=1.2.3+2.3.3+3.4.3+...+n.(n+1).3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n.(n+1)(n+2)-(n-1).n.(n+2)

=n.(n+1)(n+2)-0

=n.(n+1)(n+2)

=>A=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

=>1-1/2+1/2-1/3+...+1/x-1/(x+1)=2022/2021

=>1-1/(x+1)=2022/2021

=>1/(x+1)=-1/2021=1/-2021

=>x+1=-2021

=>x=-2022

đặt A=1/1.2+1/2.3+1/3.4+..........1/49.50

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}<1\)

vậy A<1

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

1 - 1/50 < 1

1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/49.50

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/49 - 1/50

1 - 1/50 < 1

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{10-1}{10}=\dfrac{9}{10}\)

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=2-1/1.2+3-2/2.3+4-3/3.4+...+10-9/9.10

=1-1/2+1/2-1/3+1/3-1/4+....+1/9-1/10

=1-1/10

=9/10

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{100}{100}-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(1-\dfrac{1}{x+1}=\dfrac{14}{15}\)

\(\dfrac{x+1-1}{x+1}=\dfrac{14}{15}\)

\(\dfrac{x}{x+1}=\dfrac{14}{15}\)

\(15x=14x+14\)

\(x=14\)

3S= 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) 
=>S 

Nguyễn Đình Phương giống cn gái tek?

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}< 1\) (đpcm)

ta có :

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}< 1\)

thanks mấy bạn nhiều lắm ^^