Question

TINH 1.2+2.3+3.4+......+N(N+1)

Answer 1

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 + 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)

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

Answer 2

A=1.2.3+2.3.3+3.4.3+.....+N(N+1).3

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

3A=1.2.3-1.2.0-2.3.4-2.3.1+......+N(N-1)+(N+2)-N(N-1)-N-1

3A=N(N-1)+(N+2)/3

Answer 3

Đặt S = 1.2 + 2.3 + 3.4 + .... + N( N + 1 )

3S = 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 - 1 )N( N + 1 )

=> N( N + 1 )( N + 2 )

=> S = \(\frac{N\left(N+1\right)\left(N+2\right)}{3}\)