Question

n(n+1)(n+2)(n+3)+1

Answer 1

把n(n+1)(n+2)(n+3)+1算出来,结果是n^4+6n^3+11n^2+6n+1
设
(n2+bn+c)的平方等于上面那个数
则
c^2=1
2b=6
2c+b^2=11
2bc=6
解得
b=3
c=1
所以(n^2+3n+1)^2=n(n+1)(n+2)(n+3)+1

Answer 2

把(n+1)(n+2)(n+3)(n+4)+1算出来,结果是n^4+10n^3+35n^2+50n+25
设
(n2+bn+c)的平方等于上面那个数
则
c^2=25
2b=10
2c+b^2=35
解得
b=5
c=5
所以是n^2+5n+5

Answer 3

原式=n(n+3)(n+1)(n+2)+1
=(n^2+3n)(n^2+3n+2)+1
设n^2+3n=a
所以原式=a(a+2)+1
=a^2+2a+1
=(a+1)^2
=(n^2+3n+1)^2
所以原式一定是一个完全平方式
o(∩_∩)o...