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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

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    $\begingroup$ 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. $\endgroup$
    – YiFan
    Mar 16 '20 at 3:22
  • $\begingroup$ Do you mean the first term to be ${n \choose n}$? You're using that in your attempt. $\endgroup$ Mar 16 '20 at 3:38
  • $\begingroup$ Yes. I did so that summation formula of Geometric Progression could be applied $\endgroup$
    – Maverick
    Mar 16 '20 at 3:44
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    $\begingroup$ 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.) $\endgroup$
    – joriki
    Mar 16 '20 at 4:36
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    $\begingroup$ @Robert Adding 1 shouldn't be that big of a deal... 😁 $\endgroup$
    – Wolfgang
    Mar 16 '20 at 10:28
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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.

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