N+(N+1)(N+2)(N+3)+1是哪个数的平方 快

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N+(N+1)(N+2)(N+3)+1是哪个数的平方 快

N+(N+1)(N+2)(N+3)+1是哪个数的平方 快
N+(N+1)(N+2)(N+3)+1是哪个数的平方 快

N+(N+1)(N+2)(N+3)+1是哪个数的平方 快
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

N+(N+1)(N+2)(N+3)+1=N+2+(N+1)(N+2)(N+3)-1=(N+2)[(N+1)(N+3)]-1==(N+2)[(N+2)^2-1)]-1