Simple formula for determinant of circulant matrix built on an arithmetic sequence. Let $a$ be an arithmetic sequence:
$$a_i=a_1+\lambda(i-1),\tag1$$
and consider a $n\times n$ circulant matrix $M_{n}(a)$ "built" on rotational shifts of the sequence $a$, i.e. with elements:
$$M_{ij}=a_{1+(j−i)\operatorname{mod}n}.\tag2$$
Prove:
$$
\det M_n(a)=\frac{a_1+a_n}2(-n\lambda)^{(n-1)}.\tag3
$$

An example:
  $$
\det\begin{pmatrix}
1&2&3&4&5\\
5&1&2&3&4\\
4&5&1&2&3\\
3&4&5&1&2\\
2&3&4&5&1\\
\end{pmatrix}=\frac{1+5}2(-5)^4=1875
$$

 A: Write $a$ for $a_0$ and $\newcommand{\ze}{\zeta}\ze$ for $\exp(2\pi i/n)$. The determinant is
$$\newcommand{\la}{\lambda}D=\prod_{r=0}^{n-1}\sum_{k=0}^{n-1}\ze^{jk}(a+k\la)$$
by the usual formula. The $r=0$ summand
is $$\sum_{k=0}^{n-1}(a+k\la)=na+\frac{n(n-1)}2\la.$$
The summand for $0<r\le n-1$ is
$$\sum_{k=0}^{n-1}\ze^{rk}(a+k\la)=\la\sum_{k=0}^{n-1}k\ze^{rk}.$$
This is an arithmetic-geometric progression:
$$(1-\ze^r)^2\sum_{k=0}^{n-1}k\ze^{rk}=\ze^r-n\ze^{rn}+(n-1)\ze^{r(n+1)}
=-n(1-\ze^r).$$
Then
$$\sum_{k=0}^{n-1}\ze^{rk}(a+k\la)=-\frac{n\la}{1-\ze^r}$$
and so
$$\prod_{r=1}^{n-1}
\sum_{k=0}^{n-1}\ze^{rk}(a+k\la)=(-n\la)^{n-1}\prod_{r=1}^n
\frac1{1-\ze^r}=\frac{(-n\la)^{n-1}}{n}.$$
If you multiply this by the $r=0$ summand, you should get your formula.
A: The determinant of circulant matrix $M$  is given by the formula:
$$\det M = \prod_{\ell=0}^{n-1} f(\omega^\ell)$$
where $\omega = e^{\frac{2\pi i}{n}}$ is a primitive $n^{th}$ root of unity and
$$\begin{align}
f(x) &= \sum_{k=0}^{n-1} a_{k+1} x^k 
= \sum_{k=0}^{n-1} (a_1 + \lambda k)x^k\\
&= \left(a_1 + \lambda x\frac{d}{dx}\right)\sum_{k=0}^{n-1} x^k
= \left(a_1 + \lambda x\frac{d}{dx}\right)\frac{1 - x^n}{1-x}\\
&= a_1 \frac{1 - x^n}{1-x} + \lambda\left[\frac{-n x^n}{1 -x} + x\frac{1-x^n}{(1-x)^2}\right]
\end{align}
$$
Notice $\omega,\omega^2,\cdots,\omega^{n-1}$ are all the roots of 
$g(x) = \sum_{k=0}^{n-1} x^k = \frac{1 - x^n}{1-x}$. 
We have $x^n = 1$ at these points and
$$f(\omega^\ell) = \frac{-\lambda n}{1-\omega^\ell}\quad\text{ for }\quad 1 \le \ell \le n-1$$
As a result,
$$\prod_{\ell=1}^{n-1} f(\omega^\ell) = \prod_{\ell=1}^{n-1}\frac{-\lambda n}{1 - \omega^\ell} = \frac{(-\lambda n)^{n-1}}{g(1)} = \frac{(-\lambda n)^{n-1}}{n}$$
Together with 
$$f(1) = \sum_{k=0}^{n-1} (a_1 + \lambda k) = a_1 n + \lambda \frac{n(n-1)}{2} = n\frac{a_1 + a_n}{2}$$
We obtain
$$\det M = f(1)\prod_{\ell=1}^{n-1} f(\omega^\ell) = \frac{a_1 + a_n}{2}(-\lambda n)^{n-1}$$
