Evaluating Combination Sums

Clarifying calculations

I'm realizing that the prompt might require calculating permutations (P(n,k)) for terms increasing by 4, rather than combinations (C(n,k)), which use different notation and handle k-element subsets from n elements.

Wondering about combinations

I’m thinking through whether the problem involves combinations or permutations, and if so, which specific combinations or permutations to evaluate.

Analyzing the problem

I’m working through different interpretations of the problem, including evaluating sums of combinations with indices congruent to 3 modulo 4, and considering connections to an old Olympiad problem.

Exploring sums

I'm thinking about how sums over binomial coefficients can be isolated using roots of unity, potentially simplifying the process for higher powers.

Crafting a summation

Piecing together the sum S by leveraging the binomial theorem and exponential forms. The aim is to simplify the summation over terms where k ≡ 3 mod 4.

Revisiting the sum

I'm piecing together the sum over k ≡ 3 mod 4. By reworking the expression in terms of e and summing over different s, I reach a clearer form.

Calculating complex exponentials

I’m exploring the complex exponentials $e^{-2πi s \cdot 3 /4}$ and $e^{2πi s / 4}$, evaluating them for $s = 0, 1, 2, 3$. Each value leads to specific expressions for $n$.

Breaking down the series

I’m thinking through the series term, dissecting (1 + i)^n and (1 - i)^n, and determining their real and imaginary parts. This helps streamline the calculations.

Simplifying the expression

I’m working through the exponential form of complex numbers to streamline $S$. This involves leveraging trigonometric identities and reducing powers of 2, aiming for a more concise representation.

Reworking the expression

I’m thinking about expressing S in different ways, including factoring out 2^{n/2} and simplifying the numerator. This iterative approach helps in refining the formula.

Breaking down the pattern

OK, let me see. For n = 4k, the sum of C(n,k) over k ≡ 3 mod 4 simplifies to 2^{n-2}. For n = 4k + 1, the expression includes 2^{n-2} and 2^{(n-1)/2 + 1}.

To evaluate the sum $S = \sum_{k\equiv 3 \mod 4} C(n, k)$, where $C(n, k)$ denotes the number of combinations of $n$ items taken $k$ at a time, we can utilize the properties of roots of unity and generating functions.

Step 1: Understanding the Sum

The sum involves all combinations where $k$ is congruent to 3 modulo 4. That is, $k = 3, 7, 11, \ldots$ up to $n$.

Step 2: Using Roots of Unity

We can express the sum using the discrete Fourier transform over the cyclic group of order 4. Specifically, we can use the identity involving roots of unity:

$$\sum_{k=0}^{n} C(n, k) e^{2\pi i k r/m} = (1 + e^{2\pi i r/m})^n$$

For our case, $m = 4$ and $r = 3$. Thus, $e^{2\pi i k \cdot 3/4} = e^{(3\pi i k)/2}$.

Step 3: Calculating the Generating Function

Compute $S$ using the generating function:

$$S = \frac{1}{4} \sum_{s=0}^{3} e^{-2\pi i s \cdot 3/4} (1 + e^{2\pi i s/4})^n$$

Compute each term:

1. $s = 0$:

$$e^{-2\pi i \cdot 0} = 1, \quad (1 + e^{0})^n = 2^n$$

2. $s = 1$:

$$e^{-2\pi i \cdot 3/4} = e^{-3\pi i/2} = i, \quad (1 + e^{\pi i/2})^n = (1 + i)^n$$

3. $s = 2$:

$$e^{-2\pi i \cdot 3/2} = e^{-3\pi i} = -1, \quad (1 + e^{\pi i})^n = 0^n = 0$$

4. $s = 3$:

$$e^{-2\pi i \cdot 9/4} = e^{-9\pi i/2} = -i, \quad (1 + e^{3\pi i/2})^n = (1 - i)^n$$

Step 4: Simplifying the Sum

The sum becomes:

$$S = \frac{1}{4} \left[ 1 \cdot 2^n + i (1 + i)^n + (-1) \cdot 0 + (-i) (1 - i)^n \right]$$

Simplify $(1 + i)^n$ and $(1 - i)^n$:

• Note that $1 + i = \sqrt{2} \, e^{i\pi/4}$
• Thus, $(1 + i)^n = (\sqrt{2})^n e^{i n \pi/4}$
• Similarly, $(1 - i)^n = (\sqrt{2})^n e^{-i n \pi/4}$

Compute the difference:

$$(1 + i)^n - (1 - i)^n = 2 i (\sqrt{2})^n \sin\left( \frac{n\pi}{4} \right)$$

Step 5: Final Simplification

Substitute back into the sum:

$$S = \frac{1}{4} \left[ 2^n + i \left( 2 i (\sqrt{2})^n \sin\left( \frac{n\pi}{4} \right) \right) \right] = \frac{1}{4} \left[ 2^n - 2 (\sqrt{2})^n \sin\left( \frac{n\pi}{4} \right) \right]$$

Simplify $(\sqrt{2})^n = 2^{n/2}$:

$$S = \frac{2^n - 2^{1 + n/2} \sin\left( \frac{n\pi}{4} \right)}{4}$$

Answer: An explicit expression: S = [ 2ⁿ  –  2^(1 + n⁄2) × sin(n·π⁄4) ] divided by 4; that is,

 S = [2ⁿ – 2^(1 + n⁄2)·sin(nπ⁄4)]⁄4