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From the multisections of sums section in wiki page on binomials, I found the following identity where for $t, s$, $0 \le t \lt s$

$$\tag{Ramus' identity} \sum_{k}{\binom{n}{t + ks}} = \frac{1}{s} \sum_{j=0}^{s-1}{\left(2 cos\left(\frac{j \pi}{s}\right) \right)^n} cos\left(\frac{(n - 2t)j \pi }{s}\right)$$

This has raised the question of whether $\binom{n}{0} + \binom{n}{2} + \binom{n}{4}$ can be rewritten with trig functions?


Letting $t = 0$ and $s = 2$ the left hand side of the equation expands to

$$\sum_{k}{\binom{n}{2k}} = \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \binom{n}{6} + \binom{n}{8} + \dots$$

Which is close, but there are a lot of trailing terms that are hard to get rid of. As $t \lt s$ we can't set $t = 6$, $s = 2$ and be done with it.

The only way I can see around this is by abusing the fact $\binom{n}{r} = 0$ if $r \gt n$. So by setting $s = n+1$ and $t = 6, 8, 10, 12, \dots, N$ in the above formula we can get each term. I don't know what value $N$, but a quick guess puts it as $\frac{n}{2}$ or $\frac{n-1}{2}$. We can always pick an $N$ s.t. $N \le n$.

$$t = 6 \implies \binom{n}{6} = \sum_{k}{\binom{n}{6 + k(n+1)}}$$

$$t = 8 \implies \binom{n}{8} = \sum_{k}{\binom{n}{8 + k(n+1)}}$$

$$t = 10 \implies \binom{n}{10} = \sum_{k}{\binom{n}{10 + k(n+1)}}$$

$$\vdots$$

Summing these up (with $t = 2x + 6$, and as above $s = n+1$) gives

$$\sum_{x=0}^{N}{\sum_{k}{\binom{n}{(2x + 6) + kn}}} = \sum_{x=0}^{N}{\frac{1}{n+1} \sum_{j=0}^{n}{\left(2 cos\left(\frac{j \pi}{n+1}\right) \right)^n} cos\left(\frac{(n - 2(2x + 6))j \pi }{n+1}\right)}$$

This can be simplified a little. Taking into account

$$\forall k \ge 1, \binom{n}{(2x + 6) + k(n+1)} = 0$$

We can just set $k=0$ and drop a summation on the lhs. The rhs can be reduced a little too.

$$\tag{1} \sum_{x=0}^{N}{\binom{n}{2x + 6}} = \frac{1}{n+1} \sum_{x=0}^{N}{ \sum_{j=0}^{n}{\left(2 cos\left(\frac{j \pi}{n+1}\right) \right)^n} cos\left(\frac{(n - 4x - 12)j \pi }{n+1}\right)}$$

So equation 1 gives a messy formula for all the trailing terms that are unwanted.


Plugging $t=0$, $s=2$ into Ramus' identity like above, this time for the lhs and the rhs yields

$$\sum_{k}{\binom{n}{2k}} = \frac{1}{2} \sum_{j=0}^{2-1}{\left(2 cos\left(\frac{j \pi}{2}\right) \right)^n} cos\left(\frac{(n - 0)j \pi}{2}\right)$$

$$\tag{2} \implies \sum_{k}{\binom{n}{2k}} = \frac{1}{2} \sum_{j=0}^{1}{\left(2 cos\left(\frac{j \pi}{2}\right) \right)^n} cos\left(\frac{n j \pi}{2}\right)$$


So (1) - (2) should gives me the answer.

$$\binom{n}{0} + \binom{n}{2} + \binom{n}{4} =$$

$$\frac{1}{2} \sum_{j=0}^{1}{\left(2 cos\left(\frac{j \pi}{2}\right) \right)^n} cos\left(\frac{n j \pi}{2}\right) - \frac{1}{n+1} \sum_{x=0}^{N}{\sum_{j=0}^{n}{\left(2 cos\left(\frac{j \pi}{n+1}\right) \right)^n} cos\left(\frac{(n - 4x - 12)j \pi }{n+1}\right)}$$


Questions

  1. Have I made any mistakes with the approach so far?
  2. What is the value of $N$?
  3. How can the final equation be simplified? There are a lot of common terms.

Thank you for reading this, I know it is a long post.

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