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I am trying to solve this determinant

\begin{vmatrix} 1 & 2 & \dots & n \\ 1 & 2^3 & \dots & n^3 \\ 1 & \vdots & & \vdots \\ 1 & 2^{n-1} & \dots & n^{2n-1} \end{vmatrix}

I have tried to use the Vandermonde's determinant to solve it. I also got this determinant equals:

\begin{vmatrix} 2 & \dots & n \\ 2^3-2 & \dots & n^3-n \\ \vdots & & \vdots \\ 2^{n-1}-2^{n-3} & \dots & n^{2n-1}-n^{2n-3} \end{vmatrix}

But I can get any further.

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  • $\begingroup$ What's the pattern for the entries in the determinant, exactly? (I don't know what to make of the exponents $n-1$ and $2n-1$ in the last row.) $\endgroup$ – Hans Lundmark Nov 28 '16 at 8:05
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The general term of the matrix is $c_{i,j}=j^{2i-1}$, which is almost that of a Vandermonde matrix (this would be $j^{i-1}$. Let us call $\Delta$ the wanted determinant. Let $C_j$ be the column of the matrix we are computing the determinant. Then $$\Delta=\det\left(C_1 ,\dots,C_n\right) =2\dots n\cdot \det\left(C_1,C_2/2,\dots,C_n/n\right) =:n!\Delta',$$ where $\Delta'$ is the determinant of the matrix of general term $c'_{i,j}=\left(j^2\right)^{i-1}$, which is a Vandermonde matrix. Therefore, we get $$\Delta =n!\prod_{1\leqslant j\lt j'\leqslant n}\left(j'^2-j^2\right). $$

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