mo di em a.cach lam ma ngu thi tick bat cong thoi.ngo nhu bu
S=1.2 + 2.3 + ..... + n.(n+1)
3S = 1.2.3 + 2.3.3 + ..... + n.(n+1).3
3S = 1.2.3 + 2.3.(4-1) + ...... + n.(n + 1).[(n + 2) - (n - 1)]
3S = 1.2.3 + 2.3.4 - 1.2.3 + .... + n.(n + 1).(n + 2) - (n - 1).n.(n + 1)
3S = (1.2.3 - 1.2.3) + (2.3.4 - 2.3.4) +...... + [(n-1)n(n + 1) - (n - 1).n.(n + 1)] + n.(n + 1)(n + 2)
VẬy 3S = n.(n + 1)(n + 2)
Vậy S = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
S=1.2+2.3+3.4+...+n.(n+1)
3S=1.2.3+2.3.(4-1)+...+n.(n+1)(n+2-(n-1))
3S=1.2.3+2.3.4-1.2.3+...+n(n+1)(n+2)-(n-1)n(n+1)
3S=n(n+1)(n+2)
=> S=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)