N+(N+1)(N+2)(N+3)+1是哪个数的平方快_数学解答

N+(N+1)(N+2)(N+3)+1是哪个数的平方快

人气：436 ℃　时间：2020-07-11 04:39:38

解答

回答：N*(N+1)*(N+2)*(N+3)+1
=N*(N+3)*(N+1)*(N+2)+1
=(N^2+3N)*(N^2+3N+2)+1
=(N^2+3N)^2+2(N2+3N)+1
=(N^2+3N+1)^2
为N^2+3N+1的平方