计算:n(n+1)(n+2)(n+3)+1.
问题描述:
计算:
.
n(n+1)(n+2)(n+3)+1
答
原式=
[n(n+3)[(n+1)(n+2)]+1
=
(n2+3n)[(n2+3n)+2]+1
(n2+3n)2+2(n2+3n)+1
=
(n2+3n+1)2
=n2+3n+1.