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Everyone knows that $\sum{n \choose 2k}=\sum{n \choose 2k+1}$

My question is follwing

What is the clear form of $\sum_{k=0} (-1)^k{n \choose 2k}$ and $\sum_{k=0} (-1)^k{n \choose 2k+1}$

For example, my exercise is

"calculate a+b where a=$\sum_{k=0}^{25} (-1)^k{50 \choose 2k}$ & b=$\sum_{k=0}^{24} (-1)^k{50 \choose 2k+1}$"

I found that it is relative to number 4.

If n has a form 4k+2 then a=0 and b=$(-1)^k2^{2k+1}$ or If n has a form 4k then b=0 and a=$(-1)^k2^{2k}$

But it's just my hypothesis.(I just wrote it when n=1,2,3,...,10 and found a rule)

user128766
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    I want to know simple calculation of a and b in my example by not using any recurrence formula as well as clear forms. – user128766 Apr 28 '15 at 16:39
  • You can separate k when it is an odd and even numbers, and use a change of variable. Thus, you can use $\sum{n \choose 2k}=\sum{n \choose 2k+1}$ twice. –  Apr 28 '15 at 16:40

1 Answers1

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Start with $$ (1 + i)^{n} = \sum_{k=0}^{n} \binom{n}{k} i^{k} = \sum_{h} \binom{n}{2 h} (-1)^{h} + i \sum_{h} \binom{n}{2 h + 1} (-1)^{h}, $$ and then note that \begin{align} (1 + i)^{n} &= \sqrt{2}^{n} \cdot \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right)^{n} \\&= \sqrt{2}^{n} \cdot \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right)^{n} \\&= \sqrt{2}^{n} \cdot \left( \cos\left(n \frac{\pi}{4}\right) + i \sin\left(n \frac{\pi}{4}\right) \right). \end{align}

Note that this extends the argument for the case you mentioned $$ 0 = (1 + (-1))^{n} = \sum_{h} \binom{n}{2 h} - \sum_{h} \binom{n}{2 h + 1}. $$

The download of generatingfunctionology is to be recommended.

Andreas Caranti
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