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

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/11/19 14:02:11

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?

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……

证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n 1 + (n + 1) + n*(n + 1) + n*n + (n + 1) + 1 = 2n^2 + 3n + 3 证明(1+2/n)^n>5-2/n（n属于N+,n>=3) [3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简 2^n/n*(n+1) 化简(n+1)(n+2)(n+3) 证明：1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=3^n .(n∈N+) 计算：n(n+1)(n+2)(n+3)+1 lim[(n+3)/(n+1))]^(n-2) 【n无穷大】 (1/(n^2 n 1 ) 2/(n^2 n 2) 3/(n^2 n 3) ……n/(n^2 n n)) 当N越于无穷大当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3).. 求极限Xn=n/(n^2+1)+n/(n^2+2)+n/(n^2+3)+……+n/(n^2+n),