Question

(n+1)(n+2)(n+3)（n+4)+1=?

弱弱的问一下~
(n+1)(n+2)(n+3)（n+4)+1=?
最好告诉偶一下原因，越具体越好~
3Q~

Answer 1

(n+1)(n+2)(n+3)(n+4)+1=[(n+1)(n+4)][(n+2)(n+3)]+1
=(n^2+5n+4)(n^2+5n+6)+1
令:u=n^2+5n
则原式=(u+4)(u+6)+1
=u^2+10u+24+1
=(u+5)^2
所以:
原式=(n^2+5n+5)^2

Answer 2

{n+1}{n+2}{n+3}{n+4}+1
=[{n+1}{n+4}][{n+2}{n+3}]+1
=[{n^2+5n+4}][{n^2+5n+6}]+1
=[{n^2+5n+4}][{n^2+5n+4+2}]+1
=[{n^2+5n+4}]^2+2*[{n^2+5n+4}]*1+1
=[{n^2+5n+4}+1]^2
={n^2+5n+5}^2

Answer 3

当n为任意自然数时，代数式（n+1)(n+2)(n+3)(n+4)+1是一个完全平方式

（n+1)(n+2)(n+3)(n+4)+1
=[(n^2+5n)+4][(n^2+5n)+6]+1
=(n^2+5n)^2+10(n^2+5n)+25
=[(n^2+5n)^2+1]^2