3S=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+2.3.4-2.3+3.4.5-2.3.4+......+n(n+1)(n+2) - n(n+1)(n-1)
=n(n+1) (n+2)
=>S=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] :3
k nha