分解因式：n(n+1)(n+2)(n+3)+1 步骤1=(n^2+3n)(n^2+3n+2)+1 =(n^2+3n)^2

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/24 01:01:04

分解因式：n(n+1)(n+2)(n+3)+1 步骤1=(n^2+3n)(n^2+3n+2)+1 =(n^2+3n)^2+2(n^2+3n)+1 =(n^2+3n+1)^2
想问一下步骤1是怎么来的,具体一点.

n(n+1)(n+2)(n+3)+1
步骤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

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)化简化简(n+1)(n+2)(n+3) 2^n/n*(n+1) 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).. 化简：1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)