Binomial Coefficient Challenge 3!

Prove that

∑k≥0(n+km+2k)(2kk)(−1)kk+1=(n−1m−1)\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
4 years ago

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Comments

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

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

and let Sn=∑k≥0(n+km+2k)ak\displaystyle S_n = \sum_{k\geq 0}{\binom{n+k}{m+2k}a_k}

Then consider the generating function-

S(z)=∑n≥0Snzn=∑n≥0∑k≥0(n+km+2k)akzn\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 (n+km+2k)=(n+kn−m−k)\displaystyle \binom{n+k}{m+2k} = \binom{n+k}{n-m-k}

=∑k≥0ak∑n≥0(m+2k+1+n−m−k−1n−m−k)zn−m−kzm+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}}}

=zm∑k≥0akzk∑n≥0(m+2k+1+n−m−k−1n−m−k)zn−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 1(1−z)n=∑i≥0(i+n−1i)zi\displaystyle \frac{1}{{(1-z)}^n} = \sum_{i\geq 0}{\binom{i+n-1}{i}z^i} and since (n−t)=0\displaystyle \binom{n}{-t} = 0 for positive n,tn,t, we have -

=zm∑k≥0akzk1(1−z)m+2k+1\displaystyle = z^m \sum_{k\geq 0}{a_k z^k \frac{1}{{(1-z)}^{m+2k+1}}}

=zm(1−z)m+1∑k≥0ak(z(1−z)2)k\displaystyle = \frac{z^m}{{(1-z)}^{m+1}} \sum_{k\geq 0}{a_k {\left(\frac{z}{{(1-z)}^2}\right)}^k}

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

For the current problem ak=(−1)kk+1(2kk)=Ck\displaystyle a_k = \frac{(-1)^k}{k+1} \binom{2k}{k} = C_k, where CkC_k is the kkth Catalan number.

So,

A(z)=1+4z−12z\displaystyle A(z) = \frac{\sqrt{1+4z} -1}{2z}

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

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

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

=(n−1n−m)=(n−1m−1)\displaystyle = \binom{n-1}{n-m} = \binom{n-1}{m-1} .

QED.

Kartik Sharma - 4 years ago

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@Pi Han Goh @Mark Hennings @Ishan Singh If you are interested.

Kartik Sharma - 4 years ago

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

Ishan Singh - 4 years ago

@Pi Han Goh

Kartik Sharma - 4 years ago

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I can't do this. You should be tagging @Ishan Singh and @Mark Hennings instead...

Pi Han Goh - 4 years ago

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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.

Kartik Sharma - 4 years ago

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@Kartik Sharma – Your question here is different from your deleted problem.

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

No limitation of WolframAlpha like I pointed out.

Pi Han Goh - 4 years ago
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