原式=√[n(n+1)(n+2)(n+3)+1]-(n+1)^2 =√[(n^2+3n+2)(n^2+3n)+1]-(n+1)^2 请问这一步是如何得出来的呀?
问题描述：
原式=√[n(n+1)(n+2)(n+3)+1]-(n+1)^2 =√[(n^2+3n+2)(n^2+3n)+1]-(n+1)^2 请问这一步是如何得出来的呀?
答
n(n+1)(n+2)(n+3)=(n^2+3n)(n^2+3n+2)
答
√[n(n+1)(n+2)(n+3)+1]-(n+1)² =√[(n²+3n+2)(n²+3n)+1]-(n+1)²
n(n+1)(n+2)(n+3)=[(n+1)(n+2)][n(n+3)]=(n²+3n+2)(n²+3n)
其余项不变
答
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]-...