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I'm looking for tips on how to prove the following identity for all $n \in \mathbb{N}$.

$$\sum^{\left \lfloor{\frac{n}{2}}\right \rfloor}_{k=0}\begin{pmatrix}n\\2k\end{pmatrix}=2^{n-1}$$

I know that from the binomial theorem I can show that $$2^n=(1+1)^n=\sum^n_{k=0}\begin{pmatrix}n\\k\end{pmatrix}1^k1^{n-k}=\sum^n_{k=0}\begin{pmatrix}n\\k\end{pmatrix}$$

So how can I make a connection between these two identities? Is it also possible to make a bijection between the sets representing the left and right or count in both directions?

Kookie
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  • See also https://math.stackexchange.com/questions/611761 – Bart Michels Apr 14 '20 at 08:30
  • Thanks for the suggestion. I think that's also another good problem to consider and I will definitely learn from it, but it doesn't answer my question. – Kookie Apr 14 '20 at 08:54

1 Answers1

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Hint: try to expand $(1-1)^{n}$. The value of this is zero but also equals to odd binomial coefficient substracted by even binomial coefficient

Another hint: imagine choosing a number of people from $n$ people with condition that the number must be even, then You proceed to ask the people one by one if they want to be chosen or not ($2$ possibilities). When You get to the last people, he has no choice but to join if your group is still odd or he has no choice but to not join if your group is already even.

acat3
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  • For the first hint (algebraic), can I say that $$2^n=(1+1)^n+(1-1)^n=\sum^{n}{k=0}\begin{pmatrix}n\k\end{pmatrix}+\sum^{n}{k=0}(-1)^k\begin{pmatrix}n\k\end{pmatrix}=2\sum^{\left \lfloor{\frac{n}{2}}\right \rfloor}_{k=0}\begin{pmatrix}n\2k\end{pmatrix}$$

    So now in dividing both sides by 2, we have $$2^{n-1}=\sum^{\left \lfloor{\frac{n}{2}}\right \rfloor}_{k=0}\begin{pmatrix}n\2k\end{pmatrix}$$ and I'm done.

     – Kookie Apr 14 '20 at 08:44
  • @Kookie exactly – acat3 Apr 14 '20 at 08:50
  • Got it. Thanks.

    Now I'll have a good think about the other proof. Can I ping you later on to ask you if I'm on the right track?

     – Kookie Apr 14 '20 at 08:52
  • @Kookie sure, the second hint is combinatorial proof – acat3 Apr 14 '20 at 09:11
  • Ok so, for the combinatorial proof, I can say that the left hand side represents the number of binary n-tuples with an even number of 1s. It also represents the idea of an identically sized n-tuple where each element has two options (1 or 0). Each of the first $(n-1)$ elements have 2 options, however, since there must be an even number of elements in the tuple, the $n$th term only has one choice. If there are an even number of 1s in the first $n-1$ elements, the $n$th element must be 0. Similar reasoning goes for the inverse of this statement.

    How does it sound?

     – Kookie Apr 14 '20 at 10:35
  • Thank you so much! – Kookie Apr 14 '20 at 10:35
  • @Kookie sounds good :) – acat3 Apr 14 '20 at 11:13