Combinatorial proof that binomial coefficients are given by alternating sums of squares? A student recently asked whether there was a combinatorial proof of the following identity:

$\begin{equation*}
\sum^n_{k=1}(-1)^{n-k}k^2 = {n+1 \choose 2}.
\end{equation*}$

I was in a rush and couldn't come up with anything nice off the top of my head. Any ideas or pictures to make this clearer?
More a visual proof than combinatorial, perhaps:
$$ 5^2 - 4^2 + 3^2 -2^2 +1^2 = {6 \choose 2}$$

On the left, you have the alternating sum as an inclusion-exclusion of squares: the total sum is the number of coloured cells.
On the right, you have those L shaped shapes rearranged in the top left of a 6x6 grid.
If you think of each cell as a coordinate $(x_1,x_2)$ that gives two elements chosen from the set $\{1, 2 \cdots 6\}$, it's seen that the elements are choosen  with $ x_2 &gt; x_1$, what corresponds to a combination (no repetition, and no order).
Added: The others are well known, but, just for the sake of completeness...
$$\displaystyle \sum_{k=1}^n (-1)^{n-k} k^2 = {n+1 \choose 2} = \sum_{k=1}^n \; k = \frac{(n+1) \; n}{2}$$
By induction. The formula holds for $n=0$ because both sides vanish.

You can also increase $n$ by one:
$$LHS_n = n^2 - LHS_{n-1} = \dots $$
I had to add the last term $n^2$. However, the signs of all the previous terms were reverted because of the $(-1)^n$ factor, so I had to add a minus sign in front of $LHS_{n-1}$.

Assuming that the equation holds for $n-1$, i.e. that $LHS_{n-1}=RHS_{n-1}$, the displayed formula is
$$ \dots = n^2 - RHS_{n-1} = n^2 - (n-1)n/2 = n(n+1) / 2 = RHS_n$$
which you wanted to be proved.