Is the following matrix invertible?
$\left[ \begin{matrix} \sum\limits_{x=1}^{n}{1} & \sum\limits_{x=1}^{n}{x} & \sum\limits_{x=1}^{n}{{{x}^{2}}} & \cdots & \sum\limits_{x=1}^{n}{{{x}^{k}}} \\ \sum\limits_{x=1}^{n}{x} & \sum\limits_{x=1}^{n}{{{x}^{2}}} & \sum\limits_{x=1}^{n}{{{x}^{3}}} & \cdots & \sum\limits_{x=1}^{n}{{{x}^{k+1}}} \\ \sum\limits_{x=1}^{n}{{{x}^{2}}} & \sum\limits_{x=1}^{n}{{{x}^{3}}} & \sum\limits_{x=1}^{n}{{{x}^{4}}} & \cdots & \sum\limits_{x=1}^{n}{{{x}^{k+2}}} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ \sum\limits_{x=1}^{n}{{{x}^{k}}} & \sum\limits_{x=1}^{n}{{{x}^{k+1}}} & \sum\limits_{x=1}^{n}{{{x}^{k+2}}} & \cdots & \sum\limits_{x=1}^{n}{{{x}^{2k+2}}} \\ \end{matrix} \right]$
Hint. Let $v_x=(1,x,x^2,\ldots,x^k)^T$. Note that $v_1,v_2,\ldots,v_{k+1}$ are linearly independent, because the augmented matrix $(v_1,v_2,\ldots,v_{k+1})$ is a Vandermonde matrix. Now your matrix is $A_n=\sum_{x=1}^n v_xv_x^T$. By considering quantities of the form $u^TA_nu$, it is easy to derive a necessary and sufficient condition for $A_n$ to be invertible.