Can anyone help me find a literature mentioning the following formula. I formulated it a long time ago and I am not sure if someone had thought of it before me.

\(_kC_r=\sum_{n=1}^{k-r+1} (_{k-n}C_{r-1})\) for \(k\geq r\geq 2\)

Can anyone help me find a literature mentioning the following formula. I formulated it a long time ago and I am not sure if someone had thought of it before me.

\(_kC_r=\sum_{n=1}^{k-r+1} (_{k-n}C_{r-1})\) for \(k\geq r\geq 2\)

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TopNewest@Kishlaya Jaiswal

Have you read this formula ever before ? I'm pretty sure only you can help Mr. Frago out :)

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Yep, it's indeed the Hockey-Stick Formula

I've mentioned clearly about it in my article, so you may wish to read it.

Yeah, now I've got a nice topic for a new wiki.

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It's different from the Hockey-Stick formula.

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\[\begin{eqnarray} \sum_{n=1}^{k-r+1} {k-n \choose r-1} & = & {k-1 \choose r-1} + {k-2 \choose r-1} + \ldots + {r-1 \choose r-1} \end{eqnarray}\]

Setting \(r-1=t\), and then writing the reverse expression, gives \[\begin{eqnarray} \sum_{n=1}^{k-t} {k-n \choose t} & = & \sum_{n=0}^{k-t-1} {t+n-1 \choose t} \\ & = & {t \choose t} + {t+1 \choose t} + \ldots + {k-2 \choose t} + {k-1 \choose t} \\ & = & {k \choose t+1} \\ & = & {k \choose r} \end{eqnarray}\]

which follows directly from Hockey Stick Identity

Thanks.

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But this one clearly shows that it is the Hockey-Stick:http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial

identity#Hockey-StickIdentity.Thank you very much.

EDIT: By the way, I thought of it looking into a pentagram.

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\[\left(\frac{1}{1-x}\right)^{r+1} = \left(\frac{1}{1-x}\right)^{r}\left(\frac{1}{1-x}\right)\]

and then comparing the coefficients of \(x^{k-r}\)

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EDIT: I appreciate you looking into this.

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Thanks :)

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I see , so that's why I was able to link it with you :)

I had just used it once as an identity , more like just a formula . I am wondering what else can be there to it !!

I'll be eagerly waiting for your wiki :)

P.S. Congrats on your getting 200 Friends !!

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I understand why you have said that , thanks for replying :)

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