# 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

• You can write the fraction as $\left(1+\frac1{(1+x)^n-1}\right)\left((1+x)^{nk}-1\right)$, and then use the fact that the coefficient of $x^n$ is the $n$th derivative of this expression, evaluated at $0$, divided by $n!$. This will be rather tedious to calculate but doable given enough patience. Mar 16 '20 at 3:22
• Do you mean the first term to be ${n \choose n}$? You're using that in your attempt. Mar 16 '20 at 3:38
• Yes. I did so that summation formula of Geometric Progression could be applied Mar 16 '20 at 3:44
• Do you expect that there should be a closed form? On what basis? (It seems unlikely at first sight, and Wolfram|Alpha doesn't provide one.) Mar 16 '20 at 4:36
• @Robert Adding 1 shouldn't be that big of a deal... 😁 Mar 16 '20 at 10:28

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: $$\sum\limits_{j=2}^k\binom{jn}{n}=\sum\limits_{j=2}^k\dfrac{1}{n!}\dfrac{d^{(n}}{dx}\left( (1+x)^{jn} \right)\big|_{x=0}=\dfrac{1}{n!}\dfrac{d^{(n}}{dx}\left( \sum\limits_{j=2}^k(1+x)^{jn} \right){\huge|}_{x=0}.$$