Number of solutions of $x_1+2x_2+\cdots+kx_k=n$? Suppose $n$ be a given positive integer. Then the Diophantine equation $x=n$ has only $1$ solution. Just by inspection, I found that the Diophantine equation $x+2y=n$ has $\left\lfloor \dfrac{n}{2}+1\right\rfloor$ non-negative solutions for $(x,y).$
Also, according to this post the Diophantine equation $x+2y+3z=n$ has $\left\lfloor \dfrac{n^2}{12}+\dfrac{n}{2}+1 \right\rfloor$ non-negative solutions for $(x,y).$

Is there any closed form for the number of non-negative integer solutions for $(x_1,x_2,\cdots,x_k)$ of $$x_1+2x_2+3x_3+\cdots+kx_k=n$$ for a given $k\in\Bbb{N}$?
How can I prove these formulas rigorously?

EDIT
After a very tedious calculation I found that the equation $w+2x+3y+4z=n$ has  $\left\lfloor \dfrac{n^3}{144}+\dfrac{5n^2}{48}+\dfrac{(15+(-1)^n)n}{32}+1 \right\rfloor$ solutions.
This solution completely agree with the approximation given by Rus May.
However still I believe that we can do something more than this.
 A: As Darij Grinberg says, there's not a nice closed formula for this. There is, however, a really neat approximation via Schur's theorem in combinatorics. It goes like this.
The singularities of the generating function $\frac1{(1-t)(1-t^2)\cdots(1-t^k)}$ all lie on the unit circle in the complex plane. The partial fractions decomposition of the generating function has terms of the form $\frac\alpha{(1-x/\omega)^{1+\ell}}$, where $\alpha$ is a constant, $\omega$ is a root of unity, and $\ell$ is a natural number less than $k$. The coefficient of such a term is $\alpha\binom{n+\ell}{\ell}/\omega^\ell$, so the term with the highest multiplicity makes the greatest contribution. In this case, it is the singularity at 1 with $\ell=k-1$. Then the coefficient of $t^n$ in the generating function is approximately
\begin{eqnarray*}
[t^n]\frac1{(1-t)\cdots(1-t^k)}&=&\alpha\binom{n+k-1}{k-1}+o(n^{k-1})\\
&=&\alpha\frac{n^{k-1}}{(k-1)!}+o(n^{k-1}).
\end{eqnarray*}
To evaluate the constant $\alpha$, just multiply the generating function and the partial fractions decomposition by $(1-t)^k$ and take the limit at 1, resulting in $\alpha=1/k!$. Then, Schur's approximation to the number of solutions of $x_1+2x_2+\cdots+kx_k=n$ is
$$\frac{n^{k-1}}{(k-1)!\,k!} .$$
A: Read my answer here; the number of partitions of $n$ with largest part $k$ is the same as the number of partitions into $k$ parts, which is $p(n,k)$. Using this, the number of solutions of the equation $\sum i x_{i} = n$ is equal to :
$$p(n, 1) + p(n, 2) + \cdots + p(n, k).$$
