“n(n+1)(n+2)(n+3)+1”是哪个数的平方?
问题描述：
“n(n+1)(n+2)(n+3)+1”是哪个数的平方?
答
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(n^2+3n)+1
=(n^2+3n+1)^2
所以是n^2+3n+1的平方.