Determinants

For any $n \ge 2$ the collection of polynomials $x^{n-1},(x+1)^{n-1},(x+2)^{n-1},...,(x+n)^{n-1}$ form a set of $n+1$ polynomials in the $n$ -dimensional vector space of polynomials of degree at most $n-1$ . Thus they form a linearly dependent collection, and so there exists a collection of constants $a_0,a_1,a_2,...,a_n$ (not all zero) such that $F(x) \; \equiv \; a_0 x^{n-1} + a_1(x+1)^{n-1} + a_2(x+2)^{n-1} + ... + a_n(x+n)^{n-1} \; \equiv \; 0$ If $A(n)$ is the $(n+1)\times(n+1)$ matrix $A(n)_{i,j} \; = \; (i+j-1)^{n-1} \hspace{2cm} 1 \le i,j \le n+1$ then $A(n) \left( \begin{array}{c} a_0 \\ a_1 \\ \vdots \\ a_n \end{array} \right) \; = \; \left( \begin{array}{c} F(1) \\ F(2) \\ \vdots \\ F(n+1) \end{array} \right) \; = \; \mathbf{0}$ and hence the matrix $A(n)$ is singular, and has determinant $\boxed{0}$ . In this case, we see that $A(2017)$ has determinant $\boxed{0}$ .