$$\Delta f:={\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y^2}\tag{1}$$
$$f(x,y):= \underbrace{\sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}x^{40-2k}y^{2k}}_{\text{I assume}~(x,y)\neq(0,0)} \tag{2}$$
$$\begin{align} \Delta f&= \underbrace{\Delta\left\{\sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}x^{40-2k}y^{2k}\right\}}_{\text{What I want to evaluate.} } \\&= {\partial\over\partial x} \left({\partial\over\partial x} \left( \sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}x^{40-2k}y^{2k}\right) \right) +{\partial\over\partial y} \left({\partial\over\partial y} \left( \sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}x^{40-2k}y^{2k}\right) \right)\\&= {\partial \over\partial x} \left(\sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}y^{2k} {\partial \over\partial x} x^{40-2k} \right)+{\partial \over\partial y} \left(\sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}x^{40-2k} {\partial \over\partial y} y^{2k}\right) \\&= {\partial \over\partial x} \left(\sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}y^{2k} (40-2k) x^{39-2k} \right)+{\partial \over\partial y} \left(\sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}x^{40-2k} (2k) y^{2k-1}\right) \\&= \sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}y^{2k} (40-2k) {\partial \over\partial x}x^{39-2k} +\sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}x^{40-2k} (2k){\partial \over\partial y} y^{2k-1} \\&= \sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}y^{2k} (40-2k)(39-2k)x^{38-2k} +\sum_{k=0}^{20}(-1)^{k}\binom{40}{2k}x^{40-2k} (2k)(2k-1) y^{2k-2} \\&= \sum_{k=0}^{20}(-1)^{k} \binom{40}{2k}\left\{ y^{2k} (40-2k)(39-2k)x^{38-2k} +x^{40-2k} (2k)(2k-1) y^{2k-2} \right\} \\&= \sum_{k=0}^{20}(-1)^{k} \binom{40}{2k}x^{38-2k}y^{2k-2}\left\{ y^{2} (40-2k)(39-2k) +x^{2} (2k)(2k-1) \right\} \\&= \underbrace{\sum_{k=0}^{20}2(-1)^{k} \binom{40}{2k}x^{38-2k}y^{2k-2}\left\{ (20-k)(39-2k)y^{2} + (k)(2k-1)x^2 \right\} }_{\text{Seems I can't simplify the terms moreover} } \end{align}$$
The original problem is from the $3$rd year transfer exam of math major in the university. I am confused by the person who made this problem.
PostScript
Owing to MH.Lee, I've got many progress from the initial my post's content.
And thought that I should give up evaluating the equation temporarily since achille hui showed me the quite genius equation that seems really difficult for me so far.
This is a mean question! The answer is zero.
Our first observation is that the sum looks a lot like a binomial expansion; e.g. the expansion of $(x + y)^{40}$ would be $\sum_k {40 \choose k} x^{n-k} y^k$. This sum is similar but contains only even terms; you may or may not know that the secret to isolating the even terms of a polynomial $f(x)$ is to take $\frac{f(x) + f(-x)}{2}$. Applying that observation to $(x + y)^{40}$ gives
$$\frac{(x + y)^{40} + (x - y)^{40}}{2} = \sum_k {40 \choose 2k} x^{n-2k} y^{2k}.$$
This is still not the desired expression because there is that extra factor of $(-1)^k$. But by writing $-1 = i^2$ and hence $(-1)^k = i^{2k}$ we can absorb this into $y$, producing
$$\frac{(x + iy)^{40} + (x - iy)^{40}}{2} = \sum_k (-1)^k {40 \choose 2k} x^{n-2k} y^{2k}.$$
This is much better! Now we want to apply the Laplacian to this expression. It's a straightforward computation from here, just a few applications of the power rule, but it turns out we can do even better.
At this point you might remember a fact from your complex analysis class: a function satisfying $\Delta f = 0$ is called a harmonic function, and the real or complex part of any holomorphic function is harmonic. And for $x, y$ real the above expression is the real part of the holomorphic function $z^{40}$. So its Laplacian is zero.
More explicitly, we introduce the change of variables $z = x + iy, \bar{z} = x - iy$, so that the expression is $\frac{z^{40} + \bar{z}^{40}}{2}$. This change of variables can be paired with a corresponding change of variables on differential operators, the Wirtinger derivatives
$$\frac{\partial}{\partial z} = \frac{ \frac{\partial}{\partial x} - i \frac{\partial}{\partial y}}{2}$$ $$\frac{\partial}{\partial \bar{z}} = \frac{ \frac{\partial}{\partial x} + i \frac{\partial}{\partial y}}{2}.$$
These have been constructed to satisfy the identities $\frac{\partial}{\partial z} z = \frac{\partial}{\partial \bar{z}} z = 1$ and $\frac{\partial}{\partial z} \bar{z} = \frac{\partial}{\partial \bar{z}} z = 0$. It's a pleasant fact that for a function $f : \mathbb{C} \to \mathbb{C}$ the Cauchy-Riemann equations say exactly that $\frac{\partial f}{\partial \bar{z}} = 0$, and it's another pleasant fact that the $2$-dimensional Laplacian factors as
$$\Delta = 4 \frac{\partial}{\partial z} \frac{\partial}{\partial \bar{z}}.$$
This immediately implies that $\Delta f = 0$ for $f$ any polynomial in $z$ or any polynomial in $\bar{z}$ (since the above identities give $\frac{\partial}{\partial \bar{z}} z^n = \frac{\partial}{\partial z} \bar{z}^n = 0$ by the product rule), and more generally for any function which is either holomorphic or anti-holomorphic (meaning that $\frac{\partial f}{\partial z} = 0$). The desired conclusion follows.