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I watched a youtube video (I cannot find it anymore) however, the author showed that the following binomial equation could be factored (If I remembered it correctly.)

$$\binom{n}{n+2k}=\binom{n}{n+k}\frac{1+(-1)^k}{2}$$

I cannot see how this was derived, for example I have tried the following:

$$\frac{n!}{(n+2k)!(n-(n+2k))!}$$

But cannot see how the above works.

joe_bill.dollar
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    Probably you didn't remember the formula. If $k>0$, both ${n \choose {n+2k}}$ and ${n \choose n+k}$ are zero. – aschepler Nov 29 '22 at 02:09

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What you probably saw was $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2},$$ which holds because $$\frac{1+(-1)^k}{2}= \begin{cases} 1&\text{if $2\mid k$},\\ 0 &\text{otherwise} \end{cases}$$

RobPratt
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  • Ok I see, essentially I got the k belongs to the integers, therefore we only want to iterate over the positive integers, am I correct? Therefore, for k even we get 1 and for k odd we get 0. However, If I get something like $\binom{n}{n+pk}$ with p denoting any natural number, would this become something like $\binom{n}{n+k} \cdot \frac{(p-1)+(-1)^k}{p}$? – joe_bill.dollar Nov 29 '22 at 15:11
  • No, for general $p$, you use $p$th roots of unity. See [summation] [roots-of-unity] – RobPratt Nov 29 '22 at 16:58
  • For an example with $\sum_k \binom{n}{3k+1}$, see https://math.stackexchange.com/questions/3644225/what-is-the-value-of-binomn1-binomn4-binomn7-binomn10-bi/3644277#3644277 – RobPratt Nov 29 '22 at 17:00