Câu 1.[tex]A=1.2+2.3+3.4+...+n(n+1)\Rightarrow 3A= 1.2.3+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+2)n(n+1)\Rightarrow A=\frac{n(n+1)(n+2)}{3}[/tex]
Câu 2.[tex]B=1.2.3+2.3.4+...+(n-1)n(n+1)\Rightarrow 4B=1.2.3.4+2.3.4.(5-1)+...+(n-1)n(n+1)(n+2-(n-2))=1.2.3.4+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)\Rightarrow B=\frac{(n-1)n(n+1)(n+2)}{4}[/tex]