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I'm working on a presentation on the Binomial Theorem for my Algebra 2 class and while writing Pascal's Triangle, I came across one of the properties that I haven't seen in a while. That being $$\sum_{k=1, k-j\ge0}^n\binom{k}{k-j}=\binom{n+1}{n-j}$$ I tried playing around with a proof and I'm having trouble because of the arbitrary value for $j$. I also have tried to simplify it so that I don't need the condition under the summation, that being $k-j\ge0$. For example, for $j=1, n=3$ $$1$$$$\color{red}{1} . . .1$$$$1...\color{red}{2}...1$$$$1...3...\color{red}{3}...1$$$$1...4...\color{green}{6}...4...1$$ $$\color{red}{\binom{1}{0}}+\color{red}{\binom{2}{1}}+\color{red}{\binom{3}{2}}=\binom{3+1}{3-1}=\color{green}{\binom{4}{2}}$$ However, if say $j=3$, then the first terms are negative, hence undefined, hence the need for the restriction. So is there are way to get rid of the restriction for a generalized formula?

EDIT: and no, the christmas tree theme was not intentional...but it is festive...

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  • $\begingroup$ Consider them as zeros. Again from the Pascal triangle assume that it's surrounded by zeros, the formulas will be still valid. In your example $\binom{4}{0} = \binom{3}{0} + \binom{3}{-1}+\binom{3}{-2}$ $\endgroup$ – karakfa Dec 9 '13 at 16:00
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$$\sum_{k=1, k-j\ge0}^n\binom{k}{k-j}=\binom{n+1}{n-j} \iff \sum_{k=j}^n \binom{k}{j}= \binom{n+1}{j+1} $$

This last identity can be shown using $\displaystyle {n\choose r}={n-1\choose r-1}+{n-1\choose r} $ applied repeatedly:

\begin{align} LHS &= {j\choose j}+{j+1\choose j}+{j+2\choose j}+\cdots+{n \choose j} \\ &= {j+1\choose j+1}+{j+1\choose j}+{j+2\choose j}+\cdots+{n \choose j} \\ &= {j+2\choose j+1}+{j+2\choose j}+\cdots+{n \choose j} \\ & \cdots \cdots \cdots \cdots\\ &= {n+1 \choose j+1} \end{align}

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  • $\begingroup$ very nice. I've seen that identity as well and 1) never took the time to understand it in regards to the triangle 2) should have let the restriction be $k=j$. Thanks for the help! $\endgroup$ – Eleven-Eleven Dec 9 '13 at 16:19
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    $\begingroup$ You're welcome. It is called Hockeystick identity IIRC because of the shape the coloured numbers make on the Pascal's triangle $\endgroup$ – Macavity Dec 9 '13 at 16:20
  • $\begingroup$ Ha...that's awesome. I've never heard it called that and now I'll never forget. I think my kids will like the idea (the theory and notation is way above them). $\endgroup$ – Eleven-Eleven Dec 9 '13 at 16:26

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