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 -

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.

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

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

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## Comments

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TopNewestAs 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.

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

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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 \(\dbinom{2k}{k}\) and yet another using Vandermond's identity. I'll look into it when I get time.

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@Pi Han Goh

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

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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.

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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.

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