證明連續六正整數立方和=(n^2+5n+15)(6n+15)

Question

證明連續6個正整數立方和n3+(n+1)3+(n+2)3+(n+3)3+(n+4)3+(n+5)3=(n2+5n+15)(6n+15)不只一個方法喔！寫越多(且正確)的人我選它為最佳解答。

Answer 1

1. 直接拆開  n3+(n+1)3+(n+2)3+(n+3)3+(n+4)3+(n+5)3 =n3+n3+3n2+3n+1+n3+6n2+12n+8+n3+9n2+27n+27+n2+12n2+48n+64+n3+15n2+75n+125 =6n3+45n2+165n+225 =(6n3+45n2+75n)+(90n+225) =(6n+15)(n+5)n+15(6n+15) =(6n+15)(n2+5n+15)2. 分組  n3+(n+1)3+(n+2)3+(n+3)3+(n+4)3+(n+5)3 =[n3+(n+5)3]+[(n+1)3+(n+4)3]+[(n+2)3+(n+3)3] =[(2n+5)3-3n(n+5)(2n+5)]+[(2n+5)3-3(n+1)(n+4)(2n+5)]    +[(2n+5)3-3(n+2)(n+3)(2n+5)] =(2n+5)[(2n+5)2-3n(n+5)+(2n+5)2-3(n+1)(n+4)+(2n+5)2-3(n+2)(n+3)] =(2n+5)[3(2n+5)2-3(n2+5n)-3(n2+5n+4)-3(n2+5n+6)] =3(2n+5)[(2n+5)2-(n2+5n)-(n2+5n+4)-(n2+5n+6)] =3(2n+5)[4n2+20n+25-3n2-15n-10] =3(2n+5)(n2+5n+15)3. 使用立方和公式 Σk=1n k3=13+23+33+...+n3=n2(n+1)2/4  n3+(n+1)3+(n+2)3+(n+3)3+(n+4)3+(n+5)3 =Σk=1n+5 k3-Σk=1n-1 k3 =(n+5)2(n+6)2/4-(n-1)2n2/4 =[(n+5)(n+6)+(n-1)n][(n+5)(n+6)-(n-1)n]/4 =(2n2+10n+30)(12n+30)/4 =(n2+5n+15)(6n+15)

Answer 2

還有數學歸納法

Answer 3

利用A*3+B*3=(A+B)(A*2-AB+B*2)
N*3+(N+5)*3=(2N+5)(N*2+5N+25)
(N+1)*3+(N+4)*3=(2N+5)(N*2+5N+13)
(N+2)*3+(N+3)*3=(2N+5)(N*2+5N+7)
提出公因式(2N+5)即可得