1*2*3*4+1=25=5平方2*3*4*5+1=121=11的平方3*4*5*6+1=361=19的平方

2024-12-15 02:03:59
推荐回答(6个)
回答1:

猜想n(n+1)(n+2)(n+3)+1是完全平方数
证明:
n(n+1)(n+2)(n+3)+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
所以成立。

回答2:

(n+1)(n+2)(n+3)(n+4)+1
=(n+1)(n+4)(n+2)(n+3)+1
=(n²+5n+4)(n²+5n+4+2)+1
=(n²+5n+4)²+2(n²+5n+4)+1
=(n²+5n+4+1)²
=(n²+5n+5)²

回答3:

n(n+1)(n+2)(n+3)+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
=(n^2+3n+1)^2
同理
(n+1)(n+2)(n+3)(n+4)+1=(n^2+5n+5)^2

回答4:

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

回答5:

n(n+1)(n+2)(n+3)+1=(n²+3n+1)²

回答6:

每个数的平方都是中间两个数的乘积减一
比如:3*4*5*6+1=361=19
4*5-1=19