Question

Cho A=1.2.3+2.3.4+...+n.(n+1).(n+2)

CMR:4A+1 là số chính phương 

Answer 1

 A=1.2.3+2.3.4+...+n.(n+1).(n+2)

=>4A=1.2.3.4+2.3.4.4+n(n+1)(n+2).4

=1.2.3.(4-0)+2.3.4.(5-1)+...+n.(n+1)(n+2)[(n+3)-(n-1)]

=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+n(n+1)(n+2)(n+3)-(n-1)-n.(n+1).(n+2).(n+3)

=n.(n+1)(n+2)(n+3)

=>4A+1=n(n+1)(n+2)(n+3)+1

=n.(n+3).(n+1)(n+2)+1

=(n²+3n).[n.(n+2)+1.(n+2)]+1

=(n²+3n).(n²+2n+n+2)+1

=(n²+3n).(n²+3n+2)+1

Đặt y=n²+3n

=>4A+1=y.(y+2)+1

=y²+2y+1

=y²+y+y+1

=y.(y+1)+(y+1)

=(y+1)(y+1)

=(y+1)²

Vậy 4A+1 là số chính phương