原式=√[n(n+1)(n+2)(n+3)+1]-(n+1)^2 =√[(n^2+3n+2)(n^2+3n)+1]-(n+1)^2_数学解答

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?

人气：493 ℃　时间：2020-06-15 03:21:46

解答

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]-...