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I've come across the following formula:

$$ S(n) \;:= \sum _{ \begin{array}{c} k=0 \\ k \equiv_4 0,\; 2 \end{array} } ^{ n } 2 \cdot \binom{n}{k}\cdot (-1)^{k \over 2} $$

where the summation run over the values of $k$ in $[0, n]$ that are congruent modulo 4 to 0 or 2.

Can this be simplified?


I've numerically verified that the sum takes the following values: $$ S(n) = \begin{cases} (-1)^{n \over 4} \cdot 2^{{n\over 2} + 1}, & \text{if $n \equiv_4 0 $} \\ S(n-1) , & \text{if $n \equiv_4 1 $} \\ 0, & \text{if $n \equiv_4 2 $} \\ (-1)^{n +1 \over 4} \cdot 2^{n+1\over2}, & \text{if $n \equiv_4 3 $} \\ \end{cases} $$

but except for the case $n \equiv_4 2 $ I can see why this is the case.

Particularly interesting, to me, are the fact that this sum is either 0 or a power of two and the case $n \equiv_4 1 $ where the two partial sums of the binomial coefficients in adjacent rows total the same (e.g. $ 2 \cdot 1 - 2 \cdot 6 + 2 \cdot 1 = 2 \cdot 1 - 2 \cdot 10 + 2 \cdot 5$ for the 5th and 6th row of the Pascal's triangle).

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    $\begingroup$ Two possible simplifications. The sum is over all even $k$; no need to think mod $4$ about the index. You can factor $2$ from the sum. $\endgroup$ – Ethan Bolker Jan 9 at 17:42
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$$S(n)=2\sum_{k\equiv_40,2}^n(-1)^{k/2}\binom nk=2\Big(\binom n0-\binom n2+\binom n4...\Big)$$

Consider $(1+x)^n=\binom n0+\binom n1x+\binom n2x^2+...+\binom nnx^n$.

Substitute $x=i$ to get$$(1+i)^n=\binom n0+\binom n1i-\binom n2-\binom n3i+\binom n4...$$

This gives$$\mathfrak R((1+i)^n)=\frac{S(n)}2\\\therefore S(n)=2\mathfrak R(2^{n/2}e^{in\pi/4})=2^{n/2+1}\cos(n\pi/4)$$

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