Evaluate ∑ (-x)^k ({2n Choose 2k-1} - {2n Choose 2k})

Hey guys! Today, we're diving deep into a fascinating mathematical problem, a summation that looks a bit intimidating at first glance. But don't worry, we'll break it down step by step and uncover the solution together. We're talking about evaluating the sum:

$$\sum_{k=1}^{n}(-x)^k\left[{2n \choose 2k-1} - {2n \choose 2k}\right]$$

This problem blends elements of calculus, combinatorics, and summation techniques, making it a truly engaging challenge. So, buckle up and let's get started!

The Initial Encounter: A Combinatorial Calculus Conundrum

When we first encounter this summation, the combination of the alternating signs, binomial coefficients, and the variable 'x' might seem a bit overwhelming. The key here is to recognize the underlying structures and patterns. The binomial coefficients, ${2n \choose 2k-1}$ and ${2n \choose 2k}$, immediately suggest a connection to the binomial theorem and combinatorial identities. The $(-x)^k$ term introduces an alternating sign and a power of 'x', hinting at a possible link to calculus through differentiation or integration.

Your first instinct might be to try direct computation or manipulation using standard combinatorial identities. You might recall the identity ${n \choose k} + {n \choose k-1} = {n+1 \choose k}$, which could potentially simplify the expression inside the summation. However, directly applying this identity doesn't immediately lead to a closed-form solution. Another common technique is summation by parts, which is the discrete analogue of integration by parts. This method can be useful for simplifying sums involving products of terms, but it also doesn't readily yield a solution in this case.

Instead, we need to take a step back and look for a more strategic approach. The combination of binomial coefficients and alternating signs strongly suggests exploring the binomial theorem with complex numbers. This is where the magic truly begins to happen. We'll see how expressing the sum in terms of complex exponentials unlocks a pathway to a beautiful and elegant solution. So, let’s dive deeper and see how this unfolds!

Unleashing the Power of the Binomial Theorem with Complex Numbers

The core idea to solve this summation lies in cleverly using the binomial theorem along with complex numbers. Remember the binomial theorem? It states that for any non-negative integer 'n' and any real or complex numbers 'a' and 'b':

$$(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k$$

Our strategy is to manipulate this theorem using complex numbers to isolate the terms that match our summation. Let's consider the binomial expansions of $(1 + i\sqrt{x})^{2n}$ and $(1 - i\sqrt{x})^{2n}$, where 'i' is the imaginary unit ($i^2 = -1$). Expanding these expressions using the binomial theorem, we get:

$$(1 + i\sqrt{x})^{2n} = \sum_{k=0}^{2n} {2n \choose k} (i\sqrt{x})^k$$

$$(1 - i\sqrt{x})^{2n} = \sum_{k=0}^{2n} {2n \choose k} (-i\sqrt{x})^k$$

Now, let's separate the terms in these expansions into even and odd powers of 'k'. This is crucial because our target summation involves terms with $2k-1$ and $2k$ in the binomial coefficients. By separating the even and odd terms, we can isolate these coefficients and manipulate them more effectively.

Here's where the cleverness comes in. We'll subtract the second expansion from the first. This subtraction will cause the even-powered terms (i.e., terms with $k = 2k'$) to cancel out, leaving us with only the odd-powered terms (i.e., terms with $k = 2k'-1$). Similarly, we'll consider the sum of the two expansions, which will cause the odd-powered terms to cancel out, leaving only the even-powered terms. This strategic manipulation will allow us to isolate the binomial coefficients we need for our summation.

So, let's perform these operations and see what happens. The result will be a pair of expressions that are much closer to our target summation. We'll then need to do some further algebraic manipulation to extract the desired sum. Trust me, guys, this is where the puzzle pieces start falling into place!

The Subtraction and Summation Dance: Isolating the Key Terms

Following the strategy we outlined, let's subtract the expansion of $(1 - i\sqrt{x})^{2n}$ from the expansion of $(1 + i\sqrt{x})^{2n}$:

$$(1 + i\sqrt{x})^{2n} - (1 - i\sqrt{x})^{2n} = \sum_{k=0}^{2n} {2n \choose k} (i\sqrt{x})^k - \sum_{k=0}^{2n} {2n \choose k} (-i\sqrt{x})^k$$

When we expand the sums and look closely, we'll notice that terms with even 'k' cancel out because $(i\sqrt{x})^{2k} = (-i\sqrt{x})^{2k}$. Only the terms with odd 'k' remain. Let's rewrite 'k' as $2k-1$ (where k now starts from 1) to represent the odd terms:

$$(1 + i\sqrt{x})^{2n} - (1 - i\sqrt{x})^{2n} = 2 \sum_{k=1}^{n} {2n \choose 2k-1} (i\sqrt{x})^{2k-1}$$

Now, let's simplify the term $(i\sqrt{x})^{2k-1}$. We can rewrite it as $i^{2k-1} (\,\sqrt{x})^{2k-1}$. Since $i^2 = -1$, we have $i^{2k-1} = i^{2(k-1)} * i = (-1)^{k-1}i$. Also, $(\,\sqrt{x})^{2k-1} = x^{k-1}\,\sqrt{x}$. Putting it all together:

$$(1 + i\sqrt{x})^{2n} - (1 - i\sqrt{x})^{2n} = 2i\sqrt{x} \sum_{k=1}^{n} {2n \choose 2k-1} (-x)^{k-1}$$

Next, let's add the expansions of $(1 + i\sqrt{x})^{2n}$ and $(1 - i\sqrt{x})^{2n}$:

$$(1 + i\sqrt{x})^{2n} + (1 - i\sqrt{x})^{2n} = \sum_{k=0}^{2n} {2n \choose k} (i\sqrt{x})^k + \sum_{k=0}^{2n} {2n \choose k} (-i\sqrt{x})^k$$

This time, the terms with odd 'k' cancel out, leaving only the terms with even 'k'. Let's rewrite 'k' as $2k$ (where k now starts from 0):

$$(1 + i\sqrt{x})^{2n} + (1 - i\sqrt{x})^{2n} = 2 \sum_{k=0}^{n} {2n \choose 2k} (i\sqrt{x})^{2k}$$

Simplifying $(i\sqrt{x})^{2k}$, we get $i^{2k} x^k = (-1)^k x^k$. Thus,

$$(1 + i\sqrt{x})^{2n} + (1 - i\sqrt{x})^{2n} = 2 \sum_{k=0}^{n} {2n \choose 2k} (-x)^k$$

Now we have two key expressions, one involving the sum with ${2n \choose 2k-1}$ and the other involving the sum with ${2n \choose 2k}$. We're getting closer! The next step is to combine these expressions in a way that mirrors our original summation. This will involve a simple subtraction, but the magic lies in recognizing how to manipulate these complex expressions to isolate the real part, which will give us our final answer.

The Grand Finale: Putting It All Together and Unveiling the Solution

Alright, guys, we've done the heavy lifting! We have two expressions that are tantalizingly close to our target summation. Let's recap what we've got:

$$(1 + i\sqrt{x})^{2n} - (1 - i\sqrt{x})^{2n} = 2i\sqrt{x} \sum_{k=1}^{n} {2n \choose 2k-1} (-x)^{k-1}$$

$$(1 + i\sqrt{x})^{2n} + (1 - i\sqrt{x})^{2n} = 2 \sum_{k=0}^{n} {2n \choose 2k} (-x)^k$$

Our goal is to evaluate:

$$\sum_{k=1}^{n}(-x)^k\left[{2n \choose 2k-1} - {2n \choose 2k}\right]$$

Notice the similarity between the terms inside our summation and the sums we've derived. To get the desired form, we need to manipulate our equations to isolate the difference ${2n \choose 2k-1} - {2n \choose 2k}$.

Let's start by multiplying the first equation by $-\,\sqrt{x}$:

$$- \sqrt{x} \left[ (1 + i\sqrt{x})^{2n} - (1 - i\sqrt{x})^{2n} \right] = -2ix \sum_{k=1}^{n} {2n \choose 2k-1} (-x)^{k-1}$$

Which simplifies to:

$$- \sqrt{x} \left[ (1 + i\sqrt{x})^{2n} - (1 - i\sqrt{x})^{2n} \right] = 2i \sum_{k=1}^{n} {2n \choose 2k-1} (-x)^{k}$$

Now, divide both sides by $2i$:

$$\frac{- \sqrt{x} \left[ (1 + i\sqrt{x})^{2n} - (1 - i\sqrt{x})^{2n} \right]}{2i} = \sum_{k=1}^{n} {2n \choose 2k-1} (-x)^{k}$$

Next, let's rewrite the second equation, focusing on the sum from $k=1$ to $n$ (since the $k=0$ term is simply ${2n \choose 0} = 1$):

$$(1 + i\sqrt{x})^{2n} + (1 - i\sqrt{x})^{2n} = 2 + 2 \sum_{k=1}^{n} {2n \choose 2k} (-x)^k$$

Subtracting 2 from both sides and dividing by 2, we get:

$$\frac{(1 + i\sqrt{x})^{2n} + (1 - i\sqrt{x})^{2n} - 2}{2} = \sum_{k=1}^{n} {2n \choose 2k} (-x)^k$$

Now we have expressions for both $\sum_{k=1}^{n} {2n \choose 2k-1} (-x)^{k}$ and $\sum_{k=1}^{n} {2n \choose 2k} (-x)^k$. We can finally subtract the second sum from the first to obtain our target summation:

$$\sum_{k=1}^{n}(-x)^k\left[{2n \choose 2k-1} - {2n \choose 2k}\right] = \frac{- \sqrt{x} \left[ (1 + i\sqrt{x})^{2n} - (1 - i\sqrt{x})^{2n} \right]}{2i} - \frac{(1 + i\sqrt{x})^{2n} + (1 - i\sqrt{x})^{2n} - 2}{2}$$

This looks complicated, but we're almost there! The final step is to simplify this expression. The key is to recognize that we need to isolate the real part of the expression, as our original summation is a real number. To do this, let's express $(1 + i\sqrt{x})^{2n}$ and $(1 - i\sqrt{x})^{2n}$ in polar form. Let $1 + i\sqrt{x} = r(\cos \theta + i \sin \theta)$, where $r = \sqrt{1 + x}$ and $\theta = \arctan(\sqrt{x})$. Then, $1 - i\sqrt{x} = r(\cos \theta - i \sin \theta)$.

Using De Moivre's theorem, we have:

$$(1 + i\sqrt{x})^{2n} = r^{2n} (\cos(2n\theta) + i \sin(2n\theta))$$

$$(1 - i\sqrt{x})^{2n} = r^{2n} (\cos(2n\theta) - i \sin(2n\theta))$$

Substituting these into our expression, we get:

$$\sum_{k=1}^{n}(-x)^k\left[{2n \choose 2k-1} - {2n \choose 2k}\right] = \frac{- \sqrt{x} \left[ r^{2n} (\cos(2n\theta) + i \sin(2n\theta)) - r^{2n} (\cos(2n\theta) - i \sin(2n\theta)) \right]}{2i} - \frac{r^{2n} (\cos(2n\theta) + i \sin(2n\theta)) + r^{2n} (\cos(2n\theta) - i \sin(2n\theta)) - 2}{2}$$

Simplifying further:

$$\sum_{k=1}^{n}(-x)^k\left[{2n \choose 2k-1} - {2n \choose 2k}\right] = - \sqrt{x} r^{2n} \sin(2n\theta) - r^{2n} \cos(2n\theta) + 1$$

Finally, substituting back $r = \sqrt{1 + x}$ and $\theta = \arctan(\sqrt{x})$, we arrive at the closed-form solution:

$$\sum_{k=1}^{n}(-x)^k\left[{2n \choose 2k-1} - {2n \choose 2k}\right] = -(1 + x)^n \left[ \sqrt{x} \sin(2n \arctan(\sqrt{x})) + \cos(2n \arctan(\sqrt{x})) \right] + 1$$

And there you have it! A beautiful, albeit complex, closed-form expression for our summation. This journey took us through the binomial theorem, complex numbers, De Moivre's theorem, and a good deal of algebraic manipulation. But we made it, guys! We cracked the code!

Key Takeaways and Reflections

This problem highlights the power of combining different mathematical tools to solve seemingly complex problems. The key takeaways from this exploration are:

• The binomial theorem is a versatile tool: It can be applied not just with real numbers, but also with complex numbers, leading to elegant solutions in combinatorics and summation problems.
• Complex numbers can simplify real-world problems: By using complex exponentials and De Moivre's theorem, we were able to manipulate the summation and isolate the real part, leading to our final answer.
• Strategic manipulation is crucial: The steps of subtracting and adding the binomial expansions were key to isolating the terms we needed.
• Don't be afraid to explore different approaches: Initial attempts using summation by parts might not have worked, but by thinking outside the box and using complex numbers, we found a path to the solution.

This problem serves as a great example of how mathematical beauty often lies in the connections between different areas of mathematics. By understanding these connections, we can unlock powerful techniques for solving even the most challenging problems. So, keep exploring, keep questioning, and keep the mathematical spirit alive! You never know what amazing discoveries you might make next. Until next time, happy problem-solving, guys!