原式=√[n(n+1)(n+2)(n+3)+1]-(n+1)^2 =√[(n^2+3n+2)(n^2+3n)+1]-(n

原式=√[n(n+1)(n+2)(n+3)+1]-(n+1)^2
我们把n(n+1)(n+2)(n+3)+1单独拿出来看
n(n+1)(n+2)(n+3)+1
=[n(n+3)][(n+1)(n+2)]+1
=(n^2+3n)(n^2+3n+2)+1
所以
原式=√[n(n+1)(n+2)(n+3)+1]-(n+1)^2
=√[(n^2+3n+2)(n^2+3n)+1]-(n+1)^2
一般这一类题目都可以这样考虑去做,将首尾项相乘,第二项与倒数第二项相乘……