How to compute symmetrical determinant I'm learning of determinants and am trying to find a trick to compute this one
\begin{pmatrix}
 2 & 1 & 1  & 1 & 1\\ 
 1 & 3 & 1 & 1 & 1\\ 
 1 &  1 &  4 &  1 & 1\\ 
 1 &  1 &  1 &  5 & 1\\ 
 1 &  1 &  1 &  1 & 6
\end{pmatrix}
I expanded it out and got $349$ but I feel there must be some trick to easily compute it. 

Your matrix is in a very nice form. As an alternate answer, you can use the matrix determinant lemma in this case to do the calculation quite cleanly. 
Let $\mathbf{M}$ be the matrix you provided. Then $\mathbf{M}$ = $\mathbf{A} + \mathbf{j}\mathbf{j^t}$,
where
$$
\mathbf{A} =
\begin{pmatrix}
 1 &  0 &  0 &  0 & 0\\ 
 0 &  2 &  0 &  0 & 0\\ 
 0 &  0 &  3 &  0 & 0\\ 
 0 &  0 &  0 &  4 & 0\\ 
 0 &  0 &  0 &  0 & 5
\end{pmatrix}
$$ and
$$
\mathbf{j} =
\begin{pmatrix}
  1\\ 
 1\\ 
 1\\ 
 1\\ 
 1
\end{pmatrix}.
$$
From applying the lemma, we have that $\text{det}(\mathbf{M}) = (1 + \mathbf{j^tA^{-1}j}) \ \text{det}(\mathbf{A})$.
Since $\mathbf{A}$ is diagonal in this case, the determinant is very easy to compute. It's just $5*4*3*2*1=120$. Moreover, 
$$
\mathbf{A^{-1}} =
\begin{pmatrix}
 1 &  0 &  0 &  0 & 0\\ 
 0 &  \frac{1}{2} &  0 &  0 & 0\\ 
 0 &  0 &  \frac{1}{3} &  0 & 0\\ 
 0 &  0 &  0 &  \frac{1}{4} & 0\\ 
 0 &  0 &  0 &  0 & \frac{1}{5}
\end{pmatrix}
$$ 
and since $\mathbf{j}$ is just a vector of ones, it's easy to see that 
$$\mathbf{j^tA^{-1}j} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$$.
Thus, we see that
$$ \text{det}(\mathbf{M}) = (1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}) * 120 = 394$$ 
after applying some simplification or a calculator.
