Question

试说明当N为正整数时,4N平方+12N+8的值必能写成相邻两个偶数的积

Answer 1

4n²+12n+8=4(n²+3n+2)=4(n+2)(n+1)=(2n+4)(2n+2)

y∵n为正整数

∴2n+4-2n-2=2

且(2n+4)/2=n+2

(2n+2)/2=n+1

即2n+4和2n+2为两个相邻的偶数