Question

Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

Answer 1

Ta có: \(3n\left(n+1\right)=n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

Ap dụng vào bài toán ta được

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

\(\Rightarrow3A=1.2.3+2.3.3+...+n\left(n+1\right).3\)

\(=1.2.3-0.1.2+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

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

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

Answer 2

3A = 3 [ 1.2 + 2.3 + 3.4 + ... + n(n + 1) ]

= 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 + 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 3

Ta có: A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

\(\Rightarrow\)3A = 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+2.3.4-1.2.3+3.4.5-2.3.4+...+n.(n+1).(n+2)-(n-1).n.(n+1)

3A=(n-1).n.(n+1)

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

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