1道二次根式的运算
问题描述：
1道二次根式的运算
根号N(N+1)(N+2)(N+3)+1的和
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
所以根号N(N+1)(N+2)(N+3)+1
=N^2+3N+1