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In my book it states that $\sum_{k=1}^n (\frac{C_{2k-1}^{2n-1}}{2k})=\frac{2^n-1}{n+1}$, where C is binomial coefficient. To try and prove this I did: $$\sum_{k=1}^n (\frac{C_{2k-1}^{2n-1}}{2k})=\sum_{k=0}^n (\frac{C_{2k-1}^{2n-1}}{2k})-(\frac{2n-1}{2})=\frac{1}{2n}\sum_{k=0}^n (C_{2k}^{2n})-(\frac{2n-1}{2})$$

And here the summation simplifies to $2^{2n-1}$: $$\frac{1}{2n}\sum_{k=0}^n (C_{2k}^{2n})-(\frac{2n-1}{2})=\frac{1}{2n}(2^{2n-1})-(\frac{2n-1}{2})$$ And finally it simplifies to: $$\frac{2^{2n-1}-2n^2-n}{2n}$$ I have tried every way I could think of to try and simplify this to $\frac{2^n-1}{n+1}$ but I couldn't manage it. Am I missing a way to simplify it or did I mess somewhere before? Any help is appreciated.

Edit: It turns out my assumption that the top index of the binomial coefficient was 2n-1 was incorrect, it is actually n which gives you the correct answer.

Kalpyy
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  • it might help to know that $2^n=\sum_{k=0}^n (\frac{n}{k})$ where $(\frac{n}{k})$ is the binomial coefficient – bml64 Dec 16 '23 at 15:50
  • But thats what I did right? When I expanded the sum I got $C_{0}^{2n}+C_{2}^{2n}+C_{4}^{2n}+....$ which adds up to half of $2^{2n}$ or $2^{2n-1}$ – Kalpyy Dec 16 '23 at 15:55
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    what is $\frac{C_{2k-1}}{2k}$? – D S Dec 16 '23 at 15:57
  • $C_{2k-1}$ is the binomial coefficient, or the combination formula (nCr), another representation is ${{2n-1} \choose {2k-1}}$. In my book the top part (2n-1) is not explicitly shown which is why I didn't include it here. Edit: I have now added it – Kalpyy Dec 16 '23 at 16:02
  • Does this answer your question? I came across this problem in Binomial Theorem. I need help in proving this. It seems that your guess for the missing argument was not the right one. – Anne Bauval Dec 16 '23 at 16:07
  • Yes that is true, using this does get the correct answer, but it doesn't feel right, after all how can it be n instead of 2n-1? When you put k = n you get $C_{2n-1}^n$ which means that the bottom index is larger than the top one, which I thought can't be possible? Anyways thank you for finding this, I am not sure why it didn't show up in the suggestions. – Kalpyy Dec 16 '23 at 16:15
  • When $n<k,$ $\binom nk=0$ by definition. I found this post using Approach$0$ – Anne Bauval Dec 16 '23 at 16:32

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