Trước tiên ta tính tổng
[tex]A=1.2.3+2.3.4+...+n(n+1)(n+2)[/tex]
[tex]=>4A=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+...+n(n+1)(n+2)[(n+3)-(n-1)][/tex]
[tex]=1.2.3.4+2.3.4.5-1.2.3.4 3.4.5.6-2.3.4.5+...+n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)[/tex]
[tex]=n(n+1)(n+2)(n+3)[/tex]
[tex]=>A=\frac{n(n+1)(n+2)(n+3)}{4}[/tex]
Ta tính tiếp:
[tex]B=1.2+2.3+...+n(n+1)[/tex]
[tex]=>3B=1.2.3+2.3.(4-1)+...+n.(n+1).[(n+2)-(n-1)][/tex]
[tex]=1.2.3+2.3.4-1.2.3+...+n.(n+1)(n+2)-(n-1)n(n+1)[/tex]
[tex]=n(n+1)(n+2)[/tex]
[tex]=>B=\frac{n(n+1)(n+2)}{3}[/tex]
Quay lại bài toán ta có:
[tex]S=1.2^2+2.3^2+...+n.(n+1)^2[/tex]
[tex]=1.2.(3-1)+2.3.(4-1)+...+n.(n+1)(n+2-1)[/tex]
[tex]=(1.2.3+2.3.4+...+n(n+1)(n+2)-(1.2+2.3+...+n.(n+1)[/tex]
[tex]=A-B=\frac{n.(n+1).(n+2).(n+3)}{4}-\frac{n.(n+1).(n+2)}{3}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
