一道超难的数学题:N(N+1)(N+2)(N+3)+1是哪个数的平方,并予以证明

来源:百度知道 编辑:UC知道 时间:2024/05/28 14:38:08
一道超难的数学题:N(N+1)(N+2)(N+3)+1是哪个数的平方,并予以证明。
(用完全平方知识解答)

回答: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(N2+3N)+1
=(N^2+3N+1)^2

为N^2+3N+1的平方

N(N+1)(N+2)(N+3)+1
=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

N(N+1)(N+2)(N+3)+1
=N(N+3)(N+1)(N+2)+1
=(2N+3N)(2N+3N+2)+1
=2(2N+3N)+2(2N+3N)+1
=4N+6N+4N+6N+1
=20N+1

是根号20N+1。