Question

n(n+1)(n+2)(n+3)+1 应该如何因式分解

Answer 1

n(n+1)(n+2)(n+3)+1
解：原式＝[n﹙n＋3﹚][﹙n＋1﹚﹙n＋2﹚]＋1
＝﹙n²＋3n﹚﹙n²＋3n＋2﹚＋1
设n²＋3n＝k
＝k﹙k＋2﹚＋1
＝k²＋2k＋1
＝﹙k＋1﹚²
＝﹙n²＋3n＋1﹚²

Answer 2

n(n+1)(n+2)(n+3)+1
=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2
-------------------------
^表示次方
^2表示平方
^3表示立方

Answer 3

n与n+1分解得n2+n；
n2+n再与n+2分解得n3+3n2+2n；
n3+3n2+2n再与n+3分解得n4+6n3+11n+6n；
在加1得n4+6n3+11n+6n+1.
或者n.n+1.n+2.n+3任意颠倒顺序都行！

Answer 4

n(n+1)(n+2)(n+3)+1
＝[n﹙n＋3﹚][﹙n＋1﹚﹙n＋2﹚]＋1
＝﹙n²＋3n﹚﹙n²＋3n＋2﹚＋1
=（n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2