Question

Tính A=1.2+2.3+3.4+...+n.(n+1)

giúp mk đi m.n..

Answer 1

A =1.2+2.3+3.4+.............+n(n+1)
   =1(1+1) + 2(2+1) + 3(3+1) +...+n(n+1)
   =(1^2 + 2^2 + 3^2 +...+ n^2) + (1 + 2 + 3 + ...+ n)
Ta có các công thức:
1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6
1 + 2 + 3 + ...+ n = n(n+1)/2
Thay vào ta có:
S = n(n+1)(2n+1)/6 + n(n+1)/2
=n(n+1)/2[(2n+1)/3 + 1]
=n(n+1)(n+2)/3

Answer 2

\(A=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(3A=1.2.3+2.3.4+3.4.3+..+3n\left(n+1\right)\)

\(=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(=n\left(n+1\right)\left(n+2\right)\)

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

ko chắc vì mk làm qua lâu òi hc tốt ~~:B~~

Answer 3

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 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + n.(n + 1). [(n + 2) - (n - 1)]

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

     = n . (n + 1) . (n + 2)

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

Answer 4

A = 1.2 + 2.3 + ... + n.( n + 1 ) 

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

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

3A = 12.3 + 2..4 - 1.2.3 + ... + n. (n + 1 ).( n + 2 ) - ( n - 1 ) .( n - 2 ) . ( n - 3 ) 

3A =  n .( n+ 1) . ( n + 2 ) 

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

Hc tốt

Answer 5

mk cảm ơn các bn nhiều lắm !!!!!