Question

Tính B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)

Answer 1

Ta có 4B=1.2.3.4+2.3.4.4+...+(n-1)n(n+1).4

=1.2.3.(4-0)+2.3.4.(5-1)+...+(n-1)n(n+1).(n+2-n+2)

=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1)

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

Vậy B=\(\frac{\text{(n-1)n(n+1)(n+2)}}{4}\)

Answer 2

B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)

4B = 1.2.3.4 + 2.3.4.4 + ... + (n - 1)n(n + 1).4

4B = 1.2.3.4 + 2.3.4.(5 - 1) + ... + (n - 1)n(n + 1)[(n + 2) - (n - 2)]

4B = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 + ... + (n - 1)n(n + 1)(n + 2) - (n-2)(n-1)n(n+1)

4B = (n - 1)n(n + 1)(n + 2)

B = (n - 1)n(n + 1)(n + 2) : 4

Answer 3

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

\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right).n.\left(n+1\right).4\)

\(4B=1.2.3.4+2.3.4.\left(5-1\right)+...+\left(n-1\right).n.\left(n+1\right).\)\(\left(\left(n+2\right)-\left(n-2\right)\right)\)

\(4B=\)\(1.2.3.4+2.3.4.5-1.2.3.4+...+\)\(\left(n-1\right).n.\left(n+1\right).\left(n+2\right)-\left(n-2\right).\left(n-1\right).n.\left(n+1\right)\)

\(4B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\)

=>\(B=\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)