Unlock The Fourth Term: Binomial Expansion Of (e+2f)^10

Dive Into Binomial Expansions: What Are They?

Welcome, math enthusiasts! Today, we’re going to unravel one of the coolest tools in algebra: binomial expansions. Ever wondered how you’d expand something like (a+b)^2? That’s easy, right? It's $a^2 + 2ab + b^2$. But what if you needed to expand something way more complex, like $(e+2f)^{10}$? Trying to multiply that out ten times manually would be an absolute nightmare, taking forever and practically guaranteeing mistakes! That’s precisely where the magic of binomial expansions comes into play, guys. They provide a systematic, elegant way to expand these complex algebraic expressions without the headache of endless multiplication. This isn't just a fancy math trick; it's a powerful and fundamental concept rooted in the incredible Binomial Theorem. This theorem gives us a clear, step-by-step method to expand any binomial expression (which is just an expression with two terms, like 'e' and '2f') raised to any positive integer power. It’s super useful for many reasons, especially when you only need to find a specific term within a much larger expansion, exactly like our challenge today: finding the fourth term in $(e+2f)^{10}$. Imagine you're building a huge Lego structure, and you only need one specific brick from the middle. You wouldn't empty the entire box just to find it, would you? Similarly, with the Binomial Theorem, you don't need to compute all eleven terms of $(e+2f)^{10}$ if you only care about the fourth term. The theorem allows us to pinpoint exactly that one piece of the puzzle. It simplifies this immense task, providing us with a formula that is both beautiful in its simplicity and incredibly efficient in its application. Understanding the underlying patterns and logic of how these terms are formed through repeated multiplication of a binomial is key. Each term in the expansion follows a predictable structure, combining powers of the two original terms with a special numerical factor called a binomial coefficient, which you might recognize from Pascal's Triangle or combination formulas. We’re going to break down this powerful tool together and, by the end of this guide, you'll be able to confidently tackle finding any specific term in a binomial expansion. Get ready to boost your algebraic skills and impress everyone with your newfound understanding of this fantastic mathematical concept!

The Core Formula: Deconstructing the Binomial Theorem

Alright, now that we’ve got a handle on why binomial expansions are so important and useful, let’s roll up our sleeves and dive into the heart of the matter: the formula itself! The Binomial Theorem formula is your absolute best friend when you're dealing with these expansions, especially when your goal is to locate a specific term without going through the exhaustive process of expanding the entire expression. The general formula for finding the (r+1)-th term in the expansion of $(a+b)^n$ is quite elegant: $T_{r+1} = {}_{n}C_r \cdot a^{n-r} \cdot b^r$. Let's meticulously break down each component of this powerful formula, because a solid understanding of each piece is absolutely critical for applying it correctly and avoiding common slip-ups. First up, we have n. In our problem, $(e+2f)^{10}$, n is simply the exponent, which is 10. Pretty straightforward, right? This n value indicates the total power to which the binomial is being raised, and it also tells us that there will always be n+1 terms in the full expansion. Next, we encounter a and b. These represent the two individual terms within your binomial expression. For $(e+2f)^{10}$, our a term is simply e, and our b term is 2f. Now, here's a super important alert, guys: it's absolutely crucial that you treat the entire second term, including any numerical coefficients or negative signs, as b. So, it's not just f; it's the whole 2f. This is a frequent trap that many people fall into, leading to incorrect calculations, so keep a sharp eye out for it! The next piece of the puzzle, and perhaps the most crucial for finding a specific term, is r. This little variable r is the key to determining which specific term you're actually searching for. Here’s the rule: if you’re looking for the first term, r is 0. If you need the second term, r is 1. For the third term, r is 2. Do you see the pattern emerging? Essentially, if you’re looking for the k-th term, then r will always be k-1. This is a fundamental rule of the Binomial Theorem that you simply cannot overlook! In our specific problem, we’re asked to find the fourth term, so k=4. Therefore, r will be 4 minus 1, which gives us r = 3. Got that locked in? Excellent! The term ${}_{n}C_r$ (which you might also see written as $\binom{n}{r}$) is known as the binomial coefficient. This part tells you the number of distinct ways you can choose r items from a set of n items, and it's calculated using factorials: $n! / (r! \cdot (n-r)!)$. This coefficient is what provides the unique numerical factor for each term in the expansion. Finally, we have the powers: $a^{n-r}$ and $b^r$. Notice an important relationship here: the powers of a decrease throughout the expansion, while the powers of b simultaneously increase. Also, a quick check: the sum of the exponents of a and b in any given term will always equal n. In our formula, $(n-r) + r = n$, which perfectly confirms this rule. So, for the fourth term of $(e+2f)^{10}$, we have now precisely identified n=10, a=e, b=2f, and r=3. With all these vital pieces clearly laid out, we are now perfectly equipped to substitute them into our formula and unveil that exact fourth term. Mastering this formula thoroughly is your golden ticket to conquering binomial expansions, so make sure these details are firmly etched in your mind!

Step-by-Step: Finding the Fourth Term in (e+2f)^10

Alright, guys, it's showtime! Let's take all that knowledge we've gathered and put it into action to precisely find that fourth term in the binomial expansion of $(e+2f)^{10}$. We've got our powerful formula and a clear understanding of each variable, so let's walk through this step-by-step to ensure we arrive at the correct answer without any missteps. Our general formula is $T_{r+1} = {}_{n}C_r \cdot a^{n-r} \cdot b^r$. Let's meticulously identify each component for our specific problem:

1. Identify n: Looking at our given expression, $(e+2f)^{10}$, the exponent is clearly 10. So, n = 10. This tells us the binomial is raised to the power of 10, meaning there will be 11 terms in its full expansion.

2. Identify a: The first term inside our binomial is simply e. Thus, a = e. Easy enough!

3. Identify b: This is a crucial step where careful attention is needed! The second term in the binomial is 2f. Remember, we must treat the entire term, including its coefficient, as b. So, b = 2f. Do not mistakenly use just f; that's a very common error that will lead to an incorrect result.

4. Determine r for the fourth term: As we discussed, for the k-th term, the value of r is always k-1. Since we are looking for the fourth term, k=4. Therefore, r = 4 - 1 = 3. This is arguably the most critical step; an incorrect r value will result in finding a completely different term.

Now that we have accurately identified all our variables, let's plug them directly into our general formula for the (r+1)-th term:

$T_{r+1} = {}_{n}C_r \cdot a^{n-r} \cdot b^r$

Substitute n=10, a=e, b=2f, and r=3 into the formula:

$T_{3+1} = T_4 = {}_{10}C_3 \cdot e^{(10-3)} \cdot (2f)^3$

Let’s break down each part of this exact calculation to ensure everything is correct:

• The Binomial Coefficient part: ${}_{10}C_3$. This expression means "10 choose 3," which calculates the number of ways to select 3 items from a set of 10. The formula for ${}_{n}C_r$ is $n! / (r! \cdot (n-r)!)$. So, for our term, ${}_{10}C_3 = 10! / (3! \cdot (10-3)!) = 10! / (3! \cdot 7!) = (10 \cdot 9 \cdot 8 \cdot 7!) / ((3 \cdot 2 \cdot 1) \cdot 7!)$. We can cancel out the $7!$, simplifying to $(10 \cdot 9 \cdot 8) / (3 \cdot 2 \cdot 1) = (10 \cdot 3 \cdot 4) = 120$. So, the numerical coefficient for this term is 120.

• The 'a' term part: $e^{(10-3)}$. This simplifies directly to $e^7$.

• The 'b' term part: $(2f)^3$. This is another point where precision is key. The exponent of 3 must be applied to both the numerical coefficient (2) and the variable (f) inside the parentheses. So, $(2f)^3 = 2^3 \cdot f^3 = 8f^3$. Forgetting to cube the '2' is a common error here!

Now, let's combine all these calculated pieces to form our final expression for the fourth term:

$T_4 = {}_{10}C_3 \cdot e^7 \cdot (8f^3)$

If we were to simplify this completely, it would be $120 \cdot e^7 \cdot 8f^3 = 960e^7f^3$. However, the original question asks for the expression in a format similar to the given options, specifically before the numerical multiplication of the coefficient and the b term's coefficient. Let's re-examine our derived expression in that format:

$T_4 = {}_{10}C_3 (e^7) (2f)^3$

Now, let's compare this with the provided options:

A. ${ }_{10} C_3\left(e^7\right)(2 f)^3$ – This exactly matches our meticulously derived expression!

B. ${ }_{10} C_3\left(e^7\right)(f)^3$ – Incorrect because it uses f instead of 2f for the second term.

C. ${ }_{10} C_4\left(e^6\right)(2 f)^4$ – Incorrect because it uses r=4 (which would be for the fifth term), and consequently, the exponents for e and (2f) are also wrong.

D. ${ }_{10} C_4\left(e^6\right)(f)^4$ – Incorrect for multiple reasons, combining errors seen in options B and C.

Therefore, the correct expression that represents the fourth term in the binomial expansion of $(e+2f)^{10}$ is clearly option A. See? Once you break it down into manageable steps and pay attention to those critical details, finding specific terms in binomial expansions isn't so intimidating after all. You've just mastered a powerful algebraic skill!

Common Pitfalls and Pro Tips for Binomial Expansion

Even with a clear step-by-step guide, the world of binomial expansion can still be a bit tricky, and it’s easy to stumble into some common pitfalls if you’re not vigilant. Let's talk about these potential traps and, more importantly, equip you with some pro tips to help you avoid errors and calculate terms like a seasoned math wizard! The first, and arguably most frequent, mistake we often see is the mix-up between the term number and the r value in the Binomial Theorem formula. Remember, guys, if you’re tasked with finding the k-th term, your r value in ${}_{n}C_r$ will always be k-1. So, when we were looking for the fourth term, r was 3, not 4! Many students mistakenly use r=4, which will inadvertently lead them to calculate the fifth term instead of the fourth. Always, always double-check this relationship; it’s a tiny detail that makes a huge difference in your final answer. Another major pitfall, which we emphasized earlier, is forgetting to treat the entire second term of the binomial as b. In our example, it wasn't just f; it was specifically 2f. If your b term includes a numerical coefficient (like the '2' in '2f'), a different variable, or even a negative sign (for example, if the expression was $(e - 2f)^{10}$), that entire component must be enclosed in parentheses and then raised to the power of r. Neglecting the coefficient of b, such as using f instead of 2f, will inevitably lead to an incorrect numerical factor in your final term. Similarly, when you raise that complete b term (e.g., 2f) to a specific power (e.g., 3), you must apply that exponent to every single factor inside the parentheses—both the numerical coefficient and the variable. So, $(2f)^3$ is absolutely not $2f^3$; it correctly expands to $2^3f^3$, which equals $8f^3$. Overlooking this step is a classic algebraic mistake that can easily be avoided by being meticulous with your parentheses and diligently applying exponent rules. Don't rush this part; precision here is paramount! Furthermore, while calculators can certainly help, make sure you genuinely understand the binomial coefficient notation, ${}_{n}C_r$. Knowing its manual calculation ($n! / (r! \cdot (n-r)!)$) can be a lifesaver if a calculator isn't available, and it significantly deepens your overall comprehension of how these coefficients are generated. Errors in calculating factorials or simplifying these combinations can also lead to incorrect numerical factors. Here’s a pro tip for calculating combinations manually: always look for opportunities to cancel factorials. For example, ${}_{10}C_3 = (10 \cdot 9 \cdot 8 \cdot 7!) / ((3 \cdot 2 \cdot 1) \cdot 7!) = (10 \cdot 9 \cdot 8) / (3 \cdot 2 \cdot 1) = 120$. This strategy can simplify the process considerably. Finally, and perhaps most importantly, the key to mastering any mathematical concept is practice, practice, practice! The more examples you work through, the more these patterns, rules, and exceptions will become second nature. Challenge yourself: try finding the third term or the sixth term of $(e+2f)^{10}$. What about terms in binomials with negative signs, like $(x - 3y)^7$? Or even expressions involving fractional exponents (though that's a topic for another day!). The more you engage with these types of problems, the more confident and accurate you'll become in applying the Binomial Theorem consistently and correctly every single time. By being acutely aware of these common pitfalls and consistently applying these pro tips, you'll sail through binomial expansion problems with unparalleled confidence and accuracy, sidestepping those frustrating mistakes!

Why Binomial Expansions Matter: Beyond the Classroom

At this point, you might be thinking, "Okay, I can find the fourth term now, and I understand the formula, but why does this even matter once I'm done with my math homework?" That’s a fantastic question, and the answer is that binomial expansions and the remarkable Binomial Theorem aren't just abstract mathematical concepts confined to textbooks or exam papers! They have some genuinely cool and incredibly important real-world applications across a surprisingly wide array of fields. One of the most prominent areas where you’ll consistently find binomial expansions being put to work is in the fascinating fields of probability and statistics. For instance, the binomial probability distribution, a cornerstone of statistical analysis, relies heavily on the principles of the Binomial Theorem. This distribution allows us to calculate the probability of achieving a specific number of