Cover a chessboard Let $2n\times  2n $ board. I cover it with dominoes $1\times 2$ s.t. every cell is adjacent exactly one cell coverd by a domino. I have to find the maximal number of dominoes that can be placed in this way. 
I think that the number is $\frac { n\cdot (n+1)}{2} $. I succeed for $n\in \lbrace 1, 2, 3 \rbrace $ and I try to prove it by induction. 
It's right my answer?
2 cells are adjacent  if they share a common side
 A: To show it is possible to place $n(n+1)/2$ dominoes, generalize the domino placements below.

To prove this is optimal, given a placement of dominoes, let


*

*$x$ be the number of interior dominoes,

*$y$ be the number of dominoes with both halves on the border,

*$z$ be the number of dominoes with one half on the border and the other in the interior.


Note that a perimeter domino covers $4$ perimeter squares, and a half-interior domino covers at least three perimeter squares. Since there are $4(2n-1)$ perimeter squares total, we must have
$$
4y+3z\le 4(2n-1).
$$
Furthermore, an interior domino covers $8$ squares, a border domino not on the corner covers $6$ squares, an a half-interior domino covers $7$ squares. There are at most $4$ corner dominoes, which each cover one fewer square. Therefore,
$$
8x+6y+7z\le 4n^2+4
$$
Putting this all together, we get
\begin{align}
x+y+z
&=\frac18(8x+6y+7z)+\frac1{16}(4y+7z)-\frac{5}{16}z
\\&\le \frac18(4n^2+4)+\frac1{16}(4(2n-1))
\\&= \frac12n(n+1)+\frac{1}{4}
\end{align}
Finally, since $x+y+z$ is an integer, we further conclude that $$x+y+z\le \left\lfloor\frac12n(n+1)+\frac14\right\rfloor=\frac12n(n+1).$$
