Evaluate the sum $\binom{2n}{n}+\binom{3n}{n}+\binom{4n}{n}+\cdots+\binom{kn}{n}$ Evaluate the sum $$\binom{2n}{n}+\binom{3n}{n}+\binom{4n}{n} + \cdots +\binom{kn}{n}$$
My Attempt:
Given sum = coefficient of $x^n$ in the expansion $$\{(1+x)^{n}+(1+x)^{2n}+(1+x)^{3n}+\cdots+(1+x)^{kn}\}-1 \\
= \text{coefficient of $x^n$ in}~~ \frac{(1+x)^n\{(1+x)^{nk}-1\}}{(1+x)^n-1}-1$$
But I am not able to go beyond this or there is some method using combinatorial argument
 A: The result I get is the following:
Applying the binomial expansion we have that $$(1+x)^{2n}=\sum\limits_{j=0}^{2n}\binom{2n}{j}x^j.$$ Since we want to keep track of the $n-$th coefficient, all we have to do is to derivate this function $n$ times, evaluate at $x=0$ and divide by $n!$: $$n!\binom{2n}{n}=\dfrac{d^{(n}}{dx}\left( (1+x)^{2n} \right)\big|_{x=0}\Longrightarrow \binom{2n}{n}=\dfrac{1}{n!}\dfrac{d^{(n}}{dx}\left( (1+x)^{2n} \right)\big|_{x=0}$$
Then, the sum we want to compute can be written as a sum of derivatives: $$\sum\limits_{j=2}^k\binom{jn}{n}=\sum\limits_{j=2}^k\dfrac{1}{n!}\dfrac{d^{(n}}{dx}\left( (1+x)^{jn} \right)\big|_{x=0}=\dfrac{1}{n!}\dfrac{d^{(n}}{dx}\left( \sum\limits_{j=2}^k(1+x)^{jn} \right){\huge|}_{x=0}.$$
This is the farthest I could reach when looking for a compact way of writing the initial sum. Anyways, the initial form looks way more compact and friendly. If you are looking for a method to approximate the result, applying the derivatives (numerically) could even be faster than computing the original sum. It may be worth trying.
