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Simplify $0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \dotsb,$ where $n \ge 2.$

Enter your answer in the form $f(n) 2^{g(n)},$ where $f(n)$ and $g(n)$ are polynomials in $n$ with integer coefficients.

I know how to do this question where the numbers are consecutive and I also know that $\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots = 2^{n-1}. $ How do I use this information to solve this problem?

Arctic Char
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qs13
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1 Answers1

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HINT: $k\binom{n}k=n\binom{n-1}{k-1}$

Brian M. Scott
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  • So now I have n(n-1, 0-1) + n(n-1, 2-1) + n(n-1, 4-1). (The commas between the expressions in the parenthesis represent combinations.) How do I continue? – qs13 Sep 19 '20 at 19:05
  • @132479: Factor out the $n$ and use what you already know about $\binom{n-1}1+\binom{n-1}3+\binom{n-1}5+\ldots;$. – Brian M. Scott Sep 19 '20 at 19:06
  • Is it $n2^{n-2}$? – qs13 Sep 19 '20 at 19:09
  • @132479: That’s correct: a set of $n-1$ elements has $2^{n-2}$ subsets of odd cardinality. – Brian M. Scott Sep 19 '20 at 19:13
  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Vercassivelaunos Sep 19 '20 at 19:16
  • @Vercassivelaunos: It provides a perfectly good answer, of which the OP has now made good use. – Brian M. Scott Sep 19 '20 at 19:17
  • @BrianM.Scott: I've seen such short hints given as comments, mostly, not answers. But now I see you've been here quite a bit longer than I, so you probably know better. Can I retract my recommendation to delete? – Vercassivelaunos Sep 19 '20 at 19:20
  • @Vercassivelaunos: I’ve done it both ways; in this case the hint really does most of the work, so I decided to let it stand as an answer. Then I noticed that the question had been asked and answered earlier today, so I closed the question and was about to delete my answer when the OP asked a question about it. I probably will go ahead and delete it in a bit, when I’m reasonably sure that the OP has been taken care of. – Brian M. Scott Sep 19 '20 at 19:23
  • @Vercassivelaunos: Well, I would have done so, but now that it’s been accepted, I can’t. – Brian M. Scott Sep 19 '20 at 19:28