Finding The Constant Term In (x + 1/x)^6 (x - 2/x)^5 A Step-by-Step Guide

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Hey guys! Ever stumbled upon a binomial expansion problem that made your head spin? Specifically, finding that elusive constant term? You know, the one without any 'x' hanging around? Well, you're in the right place! In this article, we're going to dive deep into how to find the constant term in the binomial expansion of expressions like (x + 1/x)^6 (x - 2/x)^5. Trust me, it's not as scary as it looks. We'll break it down step by step, making sure you've got a solid grasp on the concepts. So, grab your pencils, and let's get started!

Understanding Binomial Expansion

Before we jump into the nitty-gritty, let's quickly recap what binomial expansion is all about. Binomial expansion is basically a way of expanding expressions of the form (a + b)^n, where 'n' is a positive integer. The binomial theorem gives us a formula to do this efficiently, without having to multiply the brackets out manually (which, let's be honest, can be a real pain!). The general formula for the binomial theorem is:

(a + b)^n = Σ [nCk * a^(n-k) * b^k], where k ranges from 0 to n

Here, nCk represents the binomial coefficient, also known as "n choose k," which can be calculated as n! / (k! * (n-k)!). The exclamation mark denotes the factorial, so 5! = 5 × 4 × 3 × 2 × 1. These coefficients give us the numerical values in each term of the expansion. Think of Pascal's Triangle, where each number is the sum of the two numbers above it; the numbers in each row correspond to the binomial coefficients for a specific power 'n'. For instance, the row starting 1, 4, ... corresponds to the coefficients for (a + b)^4.

Now, in our specific problem, we are dealing with (x + 1/x)^6 (x - 2/x)^5. This looks a bit more complex, right? But don't worry, we'll tackle it systematically. The key here is to understand how each part of the expansion contributes to the final constant term. We're looking for terms where the powers of 'x' cancel each other out. This means we need to carefully consider the powers of 'x' that arise from expanding both (x + 1/x)^6 and (x - 2/x)^5.

To truly nail this, it's essential to be comfortable with manipulating exponents. Remember the rules: x^m * x^n = x^(m+n) and x^m / x^n = x^(m-n). These will be your best friends as you navigate through the expansion. We'll be using these rules to identify which terms, when multiplied together, will result in a constant (i.e., x^0). This foundational understanding of binomial expansion and exponent manipulation is crucial before we move on to the specific steps for finding the constant term in our expression. So, take a moment to ensure you're solid on these basics – they're the building blocks for everything else we'll be doing.

Breaking Down (x + 1/x)^6

Okay, let's take the first part of our expression, (x + 1/x)^6, and break it down. We'll use the binomial theorem here, but instead of expanding the whole thing, we're just interested in the general term. Why? Because we want to identify the terms that could potentially contribute to the constant term when multiplied by the expansion of (x - 2/x)^5. Remember, we're looking for terms where the 'x' powers cancel out.

Using the binomial theorem, the general term in the expansion of (x + 1/x)^6 is given by:

6Ck * x^(6-k) * (1/x)^k = 6Ck * x^(6-k) * x^(-k) = 6Ck * x^(6 - 2k)

Here, 'k' can range from 0 to 6. Notice how we simplified the expression using the exponent rules? This is crucial! Now, we have a clear view of the power of 'x' in each term: (6 - 2k). This is the key to unlocking which terms will play a role in creating a constant term.

So, what possible powers of 'x' can we get from this expansion? Let's plug in the values of k from 0 to 6:

  • k = 0: x^(6 - 2(0)) = x^6
  • k = 1: x^(6 - 2(1)) = x^4
  • k = 2: x^(6 - 2(2)) = x^2
  • k = 3: x^(6 - 2(3)) = x^0 = 1 (This is a constant term!)
  • k = 4: x^(6 - 2(4)) = x^(-2)
  • k = 5: x^(6 - 2(5)) = x^(-4)
  • k = 6: x^(6 - 2(6)) = x^(-6)

We've found a constant term within this expansion itself (when k = 3), which is 6C3 = 20. This term is just a number, with no 'x' attached. But more importantly, we've identified the powers of 'x' that can arise from this expansion: x^6, x^4, x^2, 1, x^(-2), x^(-4), and x^(-6). These are the players we need to consider when we move on to the second part of our expression. We need to see which of these can combine with terms from the expansion of (x - 2/x)^5 to give us an overall constant term. Remember, the goal is to get the exponents of 'x' to cancel each other out. Understanding these individual powers from the first binomial expansion is a critical step in solving the puzzle.

Analyzing (x - 2/x)^5

Alright, now let's shift our focus to the second binomial in our expression: (x - 2/x)^5. Just like we did with the first binomial, we're going to use the binomial theorem to find the general term. This will help us identify the powers of 'x' that we'll get from this expansion. Remember, the goal is to find terms that, when multiplied by terms from the expansion of (x + 1/x)^6, will result in a constant term (i.e., x^0).

The general term in the expansion of (x - 2/x)^5 is given by:

5Cr * x^(5-r) * (-2/x)^r = 5Cr * x^(5-r) * (-2)^r * x^(-r) = 5Cr * (-2)^r * x^(5 - 2r)

Here, 'r' can range from 0 to 5. Notice the (-2)^r term? It's crucial to keep track of the negative sign and the constant factor as we go through this. Now, let's simplify the exponent of 'x': (5 - 2r). This tells us the power of 'x' in each term of the expansion. We need to figure out which values of 'r' will give us powers of 'x' that can cancel out the powers we found in the expansion of (x + 1/x)^6.

Let's plug in the values of 'r' from 0 to 5 and see what powers of 'x' we get:

  • r = 0: x^(5 - 2(0)) = x^5
  • r = 1: x^(5 - 2(1)) = x^3
  • r = 2: x^(5 - 2(2)) = x^1 = x
  • r = 3: x^(5 - 2(3)) = x^(-1)
  • r = 4: x^(5 - 2(4)) = x^(-3)
  • r = 5: x^(5 - 2(5)) = x^(-5)

So, the possible powers of 'x' from this expansion are: x^5, x^3, x, x^(-1), x^(-3), and x^(-5). Now we have the list of 'x' powers from both expansions. The next step is to carefully pair them up to see which combinations result in x^0 (the constant term). This is where the real magic happens, and we start to see how the pieces of the puzzle fit together. Remember, the binomial coefficients (5Cr) and the (-2)^r terms will give us the numerical coefficients for each of these 'x' power terms, and we'll need to multiply those coefficients together to get the final constant term contribution from each pair.

Combining Terms to Find the Constant

Okay, this is where the real detective work begins! We've got the possible powers of 'x' from both binomial expansions: (x + 1/x)^6 and (x - 2/x)^5. Now we need to figure out which pairs of terms, when multiplied together, will give us a constant term (x^0). Remember, to get a constant term, the exponents of 'x' must add up to zero.

Let's recap the powers we have:

From (x + 1/x)^6: x^6, x^4, x^2, 1, x^(-2), x^(-4), x^(-6)

From (x - 2/x)^5: x^5, x^3, x, x^(-1), x^(-3), x^(-5)

Now, let's systematically look for pairs that add up to zero:

  1. Constant from (x + 1/x)^6 (k=3) and constant from (x - 2/x)^5: We already found that (x + 1/x)^6 has a constant term when k = 3. This term is 6C3 = 20. However, (x - 2/x)^5 does not have a constant term since 5-2r = 0 has no integer solution for r between 0 and 5. So, this case doesn't contribute to the final constant term.
  2. x^2 from (x + 1/x)^6 and x^(-2) from (x - 2/x)^5: From the first expansion, we have x^2 when k = 2. The coefficient is 6C2 = 15. From the second expansion, we need to solve 5 - 2r = -2, which gives r = 7/2, which is not an integer. Therefore, this case is not possible.
  3. x^4 from (x + 1/x)^6 and x^(-4) from (x - 2/x)^5: From the first expansion, we have x^4 when k = 1. The coefficient is 6C1 = 6. From the second expansion, we need to solve 5 - 2r = -4, which gives r = 9/2, which is not an integer. Therefore, this case is not possible.
  4. x^6 from (x + 1/x)^6 and x^(-6) from (x - 2/x)^5: Similar to the previous cases, there's no integer r between 0 and 5 that will result in x^(-6) from the second expansion. We would need to solve 5 - 2r = -6, giving r = 11/2, which is not an integer.

It seems we hit a snag. There are no direct pairs that multiply to give a constant term. This means there is no constant term in the expansion of (x + 1/x)^6 (x - 2/x)^5. So, the constant term is 0.

This might seem surprising, but it's a valuable lesson! Not all binomial expansion problems will have a constant term. Sometimes, the powers of 'x' just don't align in a way that allows them to cancel out completely. It's crucial to go through the process we've outlined here – finding the general terms, identifying the possible powers of 'x', and then carefully pairing them up – to definitively determine whether a constant term exists.

Final Calculation and Conclusion

Alright, let's wrap things up and solidify our understanding of how to find the constant term in binomial expansions like (x + 1/x)^6 (x - 2/x)^5. We've journeyed through the binomial theorem, identified general terms, and matched exponents. Now it's time for the grand finale – the final calculation.

As we meticulously paired up the powers of 'x' from the expansions of (x + 1/x)^6 and (x - 2/x)^5, we discovered something quite interesting: there were no pairs that resulted in a constant term. This means that after all the expansions and multiplications, there's no term in the final result that has x^0 (i.e., no 'x' at all).

So, what's the constant term in this case? It's simply 0. That's right, sometimes the answer is nothing! And that's perfectly okay. It highlights an important aspect of binomial expansions: not every combination of binomials will produce a constant term. The powers of 'x' need to align in a very specific way for that to happen.

To recap the process, here's what we did:

  1. We understood the binomial theorem and how it allows us to expand expressions of the form (a + b)^n.
  2. We found the general term for both (x + 1/x)^6 and (x - 2/x)^5 using the binomial theorem formula.
  3. We identified the possible powers of 'x' in each expansion by plugging in different values for 'k' and 'r' in the general terms.
  4. We paired up the powers of 'x' from both expansions, looking for combinations that would add up to zero (the condition for a constant term).
  5. We calculated the coefficients for the terms involved in the constant term, if any existed, using the binomial coefficients and any constant factors.
  6. Finally, we multiplied the coefficients to get the constant term.

In this particular problem, step 4 revealed that there were no pairs of terms that would result in a constant term. Hence, the final answer was 0.

So, what's the takeaway here, guys? Finding the constant term in binomial expansions might seem tricky at first, but with a systematic approach, it becomes much more manageable. The key is to break down the problem into smaller steps, understand the underlying principles, and carefully track your work. And remember, sometimes the answer might just be zero – and that's a perfectly valid result!

I hope this comprehensive guide has helped you demystify the process of finding constant terms in binomial expansions. Keep practicing, and you'll become a pro in no time! Happy expanding!