×

# Binomial Coefficient Challenge 3!

Prove that

$\displaystyle \sum_{k\geq 0}{\binom{n+k}{m+2k}\binom{2k}{k} \frac{{(-1)}^k}{k+1} } = \binom{n-1}{m-1}$.

Note by Kartik Sharma
2 months, 2 weeks ago

Sort by:

As there is no solution submitted, I will be posting it now.

Let's consider the generating function - $$\displaystyle A(z) = \sum_{k\geq 0}{a_k z^k}$$

and let $$\displaystyle S_n = \sum_{k\geq 0}{\binom{n+k}{m+2k}a_k}$$

Then consider the generating function-

$\displaystyle S(z) = \sum_{n\geq 0}{S_n z^n} = \sum_{n\geq 0}{\sum_{k\geq 0}{\binom{n+k}{m+2k} a_k z^n}}$

Replacing $$\displaystyle \binom{n+k}{m+2k} = \binom{n+k}{n-m-k}$$

$\displaystyle = \sum_{k\geq 0}{a_k\sum_{n\geq 0}{\binom{m+ 2k +1 + n- m -k -1}{n-m-k}z^{n-m-k}z^{m+k}}}$

$\displaystyle = z^m \sum_{k\geq 0}{a_k z^k \sum_{n\geq 0}{\binom{m+ 2k +1 + n- m -k -1}{n-m-k}z^{n-m-k}}}$

Since $$\displaystyle \frac{1}{{(1-z)}^n} = \sum_{i\geq 0}{\binom{i+n-1}{i}z^i}$$ and since $$\displaystyle \binom{n}{-t} = 0$$ for positive $$n,t$$, we have -

$\displaystyle = z^m \sum_{k\geq 0}{a_k z^k \frac{1}{{(1-z)}^{m+2k+1}}}$

$\displaystyle = \frac{z^m}{{(1-z)}^{m+1}} \sum_{k\geq 0}{a_k {\left(\frac{z}{{(1-z)}^2}\right)}^k}$

$\displaystyle S(z) = \frac{z^m}{{(1-z)}^{m+1}}A\left(\frac{z}{{(1-z)}^2}\right)$

For the current problem $$\displaystyle a_k = \frac{(-1)^k}{k+1} \binom{2k}{k} = C_k$$, where $$C_k$$ is the $$k$$th Catalan number.

So,

$\displaystyle A(z) = \frac{\sqrt{1+4z} -1}{2z}$

$\displaystyle A\left(\frac{z}{{(1-z)}^2}\right) = 1 - z$

$\displaystyle S(z) = \frac{z^m}{{(1-z)}^{m+1}} (1-z) = \frac{z^m}{{(1-z)}^m}$

We need to find $$\displaystyle S_n = [z^n]{S(z)}$$ which is

$\displaystyle = \binom{n-1}{n-m} = \binom{n-1}{m-1}$.

QED. · 2 months, 1 week ago

@Pi Han Goh @Mark Hennings @Ishan Singh If you are interested. · 2 months, 1 week ago

Nice. Snake oil is the go to solution in many Binomial Sums. I think we can have another solution using the integral representation of $$\dbinom{2k}{k}$$ and yet another using Vandermond's identity. I'll look into it when I get time. · 2 months, 1 week ago

@Pi Han Goh · 2 months, 2 weeks ago

I can't do this. You should be tagging @Ishan Singh and @Mark Hennings instead... · 2 months, 2 weeks ago

I was just saying that it was this problem which W|A previously failed to give the correct answer. Here. We have seen a limit of W|A. · 2 months, 2 weeks ago

In your deleted problem, the summation is $$\displaystyle \sum_{k=0}^m$$, whereas in your question here, it states $$\displaystyle \sum_{k\geq 0}$$.

No limitation of WolframAlpha like I pointed out. · 2 months, 2 weeks ago