A = 1.2.4 + 2.3.5 + ... + n(n+1)(n+3)
A = 1.2.(3+1) + 2.3.(4+1) + ... + n(n+1)[(n+2)+1]
A = [1.2.3 + 2.3.4 + ... + n(n+1)(n+2)] + [1.2 + 2.3 + ... + n(n+1)]
Đặt B = 1.2.3 + 2.3.4 + ... + n(n+1)(n+2)
4B = 1.2.3.(4-0) + 2.3.4.(5-1) + ... + n(n+1)(n+2)[(n+3)-(n-1)]
4B = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)
4B = n(n+1)(n+2)(n+3)
B = \(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
Đặt C = 1.2 + 2.3 + ... + n(n+1)
3C = 1.2.(3-0) + 2.3.(4-1) + ... + n(n+1)[(n+2)-(n-1)]
3C = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + ... + n(n+1)(n+2) - (n-1)n(n+1)
3C = n(n+1)(n+2)
C = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
A = B + C = \(n\left(n+1\right)\left(n+2\right)\left(\frac{n+3}{4}+\frac{1}{3}\right)\)
\(=n\left(n+1\right)\left(n+2\right)\frac{3n+13}{12}\)
