Coefficient Of X^6 In (2x-5)^9 Expansion

Unlocking the Coefficient: A Deep Dive into Binomial Expansions

When we talk about binomial expansion, we're essentially discussing a powerful mathematical technique that allows us to expand expressions of the form $(a+b)^n$ into a sum of terms. This concept is fundamental in various fields, from probability and statistics to calculus and beyond. Today, we're going to tackle a specific problem: finding the coefficient of a particular term, $x^6$, in the expansion of the binomial expression $(2x-5)^9$. This might sound a bit daunting at first, but with the right tools and a step-by-step approach, it becomes an accessible and even enjoyable puzzle to solve. The Binomial Theorem is our guiding star here. It provides a general formula for expanding any positive integer power of a binomial. The theorem states that for any non-negative integer n, the expansion of $(a+b)^n$ is given by:

(a+b)^n = inom{n}{0}a^n b^0 + inom{n}{1}a^{n-1} b^1 + inom{n}{2}a^{n-2} b^2 + ext{...} + inom{n}{k}a^{n-k} b^k + ext{...} + inom{n}{n}a^0 b^n

where inom{n}{k} represents the binomial coefficient, calculated as rac{n!}{k!(n-k)!}. This formula systematically breaks down the complex expression into simpler, manageable terms. Each term in the expansion is a product of a binomial coefficient, a power of the first term (a), and a power of the second term (b). The exponents of a decrease from n to 0, while the exponents of b increase from 0 to n, and the sum of the exponents in each term always equals n. Understanding this general form is crucial for solving our specific problem, as it gives us a blueprint for how to construct and analyze each part of the expansion. We'll be using this theorem to pinpoint the exact term that contains $x^6$ and then extract its coefficient. It's like being a detective, looking for a specific clue within a larger piece of evidence, and the Binomial Theorem is our magnifying glass.

Applying the Binomial Theorem to Our Specific Case: $(2x-5)^9$

Now, let's bring the Binomial Theorem to bear on our specific expression, $(2x-5)^9$. In this case, our a is $2x$ and our b is $-5$, and n is 9. The general term in the expansion of $(a+b)^n$ is given by inom{n}{k}a^{n-k}b^k. Substituting our values, the general term for $(2x-5)^9$ becomes inom{9}{k}(2x)^{9-k}(-5)^k. Our goal is to find the term where the power of x is 6. Looking at the general term, the power of x comes from $(2x)^{9-k}$. For the power of x to be 6, the exponent $9-k$ must be equal to 6. So, we set up the equation: $9-k = 6$. Solving for k, we get $k = 9-6 = 3$. This tells us that the term containing $x^6$ will be the term where $k=3$. Now that we know the value of k, we can substitute it back into the general term formula:

Term = inom{9}{3}(2x)^{9-3}(-5)^3

Term = inom{9}{3}(2x)^6(-5)^3

At this stage, we need to calculate the binomial coefficient inom{9}{3} and the powers of 2 and -5. The binomial coefficient inom{9}{3} is calculated as rac{9!}{3!(9-3)!} = rac{9!}{3!6!} = rac{9 imes 8 imes 7}{3 imes 2 imes 1} = 3 imes 4 imes 7 = 84. Next, we have $(2x)^6 = 2^6 imes x^6 = 64x^6$. And finally, $(-5)^3 = -125$. Now, we multiply all these parts together to get the complete term:

Term = $84 imes (64x^6) imes (-125)$

Term = $84 imes 64 imes (-125) imes x^6$

To find the coefficient, we just need to compute the product of the numerical parts: $84 imes 64 imes (-125)$. This calculation might seem tedious, but it's a straightforward multiplication. $84 imes 64 = 5376$. Then, $5376 imes (-125) = -672000$. So, the term containing $x^6$ is $-672000x^6$. The coefficient of $x^6$ is therefore $-672000$. This process highlights how the Binomial Theorem allows us to isolate and calculate specific components of a complex expansion without having to compute every single term. It's a testament to the elegance and efficiency of algebraic principles.

Calculating the Binomial Coefficient and Powers: The Nuts and Bolts

Let's delve deeper into the mechanics of calculating the components that form our target coefficient. The binomial coefficient, inom{n}{k}, is a cornerstone of the Binomial Theorem, and its accurate calculation is paramount. In our case, for $k=3$ and $n=9$, we calculated inom{9}{3} as 84. This number represents the number of ways to choose 3 items from a set of 9, without regard to the order of selection. The formula rac{n!}{k!(n-k)!} is the standard way to compute this. Remember that the factorial of a non-negative integer 'm', denoted by m!, is the product of all positive integers less than or equal to m. For example, $5! = 5 imes 4 imes 3 imes 2 imes 1 = 120$. So, $9! = 9 imes 8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1$, and $6! = 6 imes 5 imes 4 imes 3 imes 2 imes 1$. When we compute rac{9!}{3!6!}, the $6!$ in the numerator and denominator cancels out, leaving us with rac{9 imes 8 imes 7}{3 imes 2 imes 1}. This simplification makes the calculation much more manageable. The result, 84, is a critical multiplier in our final coefficient.

Equally important are the powers of the terms within the binomial. We have $(2x)^6$ and $(-5)^3$. For $(2x)^6$, we need to raise both the coefficient 2 and the variable x to the power of 6. This means $2^6 imes x^6$. Calculating $2^6$: $2 imes 2 = 4$, $4 imes 2 = 8$, $8 imes 2 = 16$, $16 imes 2 = 32$, $32 imes 2 = 64$. So, $2^6 = 64$. This gives us $64x^6$. The $x^6$ part is what we are interested in for identifying the term, while the 64 is the numerical factor associated with $2x$ in this term.

Next, we deal with $(-5)^3$. When raising a negative number to an odd power, the result is negative. So, $(-5)^3 = (-5) imes (-5) imes (-5)$. First, $(-5) imes (-5) = 25$. Then, $25 imes (-5) = -125$. The number $-125$ is the numerical value for the second part of our binomial term.

Putting it all together, the term is the product of these three calculated values: the binomial coefficient (84), the numerical part of the first term raised to its power (64), and the second term raised to its power (-125). The full numerical coefficient is $84 imes 64 imes (-125)$. As calculated before, $84 imes 64 = 5376$. Then, $5376 imes (-125)$. To perform this multiplication efficiently, one can think of multiplying by 1000/8. So, 5376 imes rac{1000}{8} = rac{5376000}{8}. Dividing 5376000 by 8: $53 ext{ divided by } 8 ext{ is } 6 ext{ with remainder } 5$. $57 ext{ divided by } 8 ext{ is } 7 ext{ with remainder } 1$. $16 ext{ divided by } 8 ext{ is } 2$. So, $5376000 / 8 = 672000$. Since we were multiplying by -125, the result is $-672000$. Thus, the coefficient of $x^6$ in the expansion of $(2x-5)^9$ is $-672000$. This detailed breakdown ensures that each component is correctly computed and combined, leading to the accurate final answer.

Conclusion: The Power of the Binomial Theorem in Action

In conclusion, we have successfully determined the coefficient of $x^6$ in the expansion of $(2x-5)^9$ by systematically applying the Binomial Theorem. We identified that the general term in the expansion is given by inom{n}{k}a^{n-k}b^k. For our specific problem, where $a=2x$, $b=-5$, and $n=9$, the general term is inom{9}{k}(2x)^{9-k}(-5)^k. To find the term with $x^6$, we equated the exponent of x to 6, meaning $9-k=6$, which yielded $k=3$. Substituting $k=3$ back into the general term formula, we obtained inom{9}{3}(2x)^6(-5)^3. We then meticulously calculated each component: the binomial coefficient inom{9}{3} = 84, the power $(2x)^6 = 64x^6$, and the power $(-5)^3 = -125$. Multiplying these numerical values together, $84 imes 64 imes (-125)$, we arrived at the final coefficient of $-672000$. This exercise underscores the immense utility of the Binomial Theorem as a precise and efficient tool for dissecting complex algebraic expressions. It not only simplifies the process of finding specific terms but also provides a robust framework for understanding polynomial expansions in general.

For further exploration into the fascinating world of binomial expansions and related mathematical concepts, you might find the resources at Wolfram MathWorld incredibly insightful. They offer comprehensive definitions, theorems, and examples that can deepen your understanding.