A special case: determinant of a $n\times n$ matrix I would like to solve for the determinant of a $n\times n$ matrix $V$ defined as:
$$
V_{i,j}=
\begin{cases}
v_{i}+v_{j} & \text{if} & i \neq j \\[2mm]
(2-\beta_{i}) v_{i} & \text{if} & i = j \\
\end{cases}
$$
Here, $v_i, v_j \in(0,1)$, and $\beta_i>0$.  Any suggestions or thoughts are highly appreciated.  

By investigating the structure of the result I come to the nice general formula
$$
\left|V\right| = (-1)^n \left(1-\sum _{i=1}^n \left(\frac{2}{\beta _i}+\sum _{j=1}^{i-1}
   \frac{\left(v_i-v_j\right){}^2}{v_i v_j \beta _i \beta _j}\right)\right)
   \prod _{k=1}^n v_k \beta _k
$$
If you have Mathematica you can use the following code to check the result
n = 4;
V = Table[If[i == j, (2 - β[i]) v[i], v[i] + v[j]], {i, n}, {j, n}];
(-1)^n (1 - Sum[2/β[i] + Sum[(v[i] - v[j])^2/(
   v[i] v[j] β[i] β[j]), {j, i - 1}], 
   {i, n}]) Product[v[k] β[k], {k, n}] == Det[V] // Expand
(* True *)

I believe that there is a proof for this formula but I don't find it yet.
