Completing Mathematical Expressions A Step By Step Guide

In the realm of mathematics, expressions often present themselves as intriguing puzzles, inviting us to unravel their hidden patterns and complete their sequences. This article delves into a specific expression, challenging us to identify the missing terms and restore its integrity. We'll explore the underlying principles governing the expression, dissect its components, and ultimately, piece together the final four terms, gaining a deeper understanding of mathematical expressions along the way.

Understanding the Expression's Foundation

Before we embark on the quest to complete the expression, it's crucial to grasp the fundamental concepts at play. Our expression, $128 x^{21}+1,344 x^{18} y^2+6,048 x^{15} y^4+15,120 x^{12} y^6+\cdots + \boxed{\phantom{0000}} x^9 y^8$, hints at a binomial expansion, a powerful tool for expanding expressions of the form $(a + b)^n$. Binomial expansions follow a predictable pattern, governed by the binomial theorem, which provides a formula for determining the coefficients and exponents of each term. Let's dissect this theorem to unveil its secrets.

The binomial theorem states that for any non-negative integer n, the expansion of $(a + b)^n$ can be expressed as a sum of terms, each with a specific coefficient and combination of a and b raised to certain powers. The coefficients are determined by binomial coefficients, often denoted as "n choose k" or $\binom{n}{k}$, which represent the number of ways to choose k items from a set of n items. These coefficients can be calculated using the formula:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

where n! represents the factorial of n, the product of all positive integers up to n. The exponents of a and b in each term follow a pattern as well. As we move from one term to the next, the exponent of a decreases by 1, while the exponent of b increases by 1. The sum of the exponents in each term always equals n, the power to which the binomial is raised.

With this foundation in place, we can begin to dissect our expression and identify the values of a, b, and n that govern its expansion. By carefully examining the existing terms, we can deduce the pattern and predict the missing terms with confidence. This process of reverse engineering, of starting with the expanded form and working backward to the original binomial, is a testament to the elegance and interconnectedness of mathematics.

Unraveling the Expression's Pattern

Now, let's turn our attention back to the expression at hand: $128 x^{21}+1,344 x^{18} y^2+6,048 x^{15} y^4+15,120 x^{12} y^6+\cdots + \boxed{\phantom{0000}} x^9 y^8$. Our mission is to decipher the pattern within the known terms and extrapolate it to fill in the blanks.

The first clue lies in the decreasing powers of x. We observe that the exponent of x starts at 21 and diminishes by 3 in each subsequent term (21, 18, 15, 12). This suggests that the x term in our binomial is being raised to a power that decreases by 3 with each term. Conversely, the exponent of y increases by 2 in each term (0, 2, 4, 6), indicating that the y term in our binomial is being raised to a power that increases by 2. This dance of exponents, one decreasing while the other increases, is a hallmark of binomial expansions.

The next key lies in the coefficients of the terms (128, 1344, 6048, 15120). These coefficients, while seemingly random at first glance, are the product of the binomial coefficients and the constants within the a and b terms of our binomial. To unlock their secret, we need to identify the values of a, b, and n that generated these coefficients.

Let's hypothesize that our binomial expression takes the form $(Ax^p + By^q)^n$, where A, B, p, q, and n are constants that we need to determine. By matching the powers of x and y in our expression with the general form of a binomial expansion, we can begin to narrow down the possibilities.

The highest power of x in our expression is 21, and the highest power of y is 8. This suggests that $x^p$ raised to the power of n might give us $x^{21}$, and $y^q$ raised to some power might contribute to the $y^8$ term. Furthermore, the exponents of x decrease by 3, hinting that p might be 3, and the exponents of y increase by 2, suggesting that q might be 2. This initial detective work lays the groundwork for a more rigorous analysis.

The Quest to Complete the Expression: Unveiling the Missing Terms

With our initial clues gathered, let's embark on the exciting phase of completing the expression. Our working hypothesis is that the expression stems from a binomial expansion of the form $(Ax^3 + By^2)^n$. To confirm this and determine the values of A, B, and n, we'll analyze the coefficients and exponents of the known terms more closely.

The first term, $128x^{21}$, gives us a valuable starting point. If our hypothesis is correct, this term corresponds to the first term in the binomial expansion, which is given by $\binom{n}{0}(Ax^3)^n(By^2)^0$. This simplifies to $A^n x^{3n}$. Comparing this to $128x^{21}$, we deduce that $3n = 21$, which means n = 7. Additionally, $A^n = A^7 = 128$, which implies that A = 2.

Now that we know n = 7 and A = 2, let's examine the second term, $1344x^{18}y^2$. This term corresponds to the second term in the binomial expansion, given by $\binom{n}{1}(Ax^3)^{n-1}(By^2)^1$, which simplifies to $n A^{n-1} B x^{3(n-1)} y^2$. Substituting n = 7 and A = 2, we get $7 imes 2^6 imes B x^{18} y^2$. Comparing this to $1344x^{18}y^2$, we have $7 imes 64 imes B = 1344$, which leads to B = 3.

With A = 2, B = 3, and n = 7, we've successfully unmasked the binomial expression: $(2x^3 + 3y^2)^7$. Now, we can confidently generate the missing terms using the binomial theorem. The general term in the expansion is given by:

$\binom{7}{k}(2x^3)^{7-k}(3y^2)^k$

where k ranges from 0 to 7. We already have the terms for k = 0, 1, 2, and 3. The next term in the given expression corresponds to k = 4, which is:

$\binom{7}{4}(2x^3)^{3}(3y^2)^4 = 35 imes 8 x^9 imes 81 y^8 = 22680 x^9 y^8$

Thus, the missing term is $22680 x^9 y^8$.

To complete the last four terms, we need to calculate the terms for k = 5, 6, and 7:

For k = 5: $\binom{7}{5}(2x^3)^{2}(3y^2)^5 = 21 imes 4 x^6 imes 243 y^{10} = 20412 x^6 y^{10}$

For k = 6: $\binom{7}{6}(2x^3)^{1}(3y^2)^6 = 7 imes 2 x^3 imes 729 y^{12} = 10206 x^3 y^{12}$

For k = 7: $\binom{7}{7}(2x^3)^{0}(3y^2)^7 = 1 imes 1 imes 2187 y^{14} = 2187 y^{14}$

Therefore, the complete last four terms of the expression are: $22680 x^9 y^8 + 20412 x^6 y^{10} + 10206 x^3 y^{12} + 2187 y^{14}$. We have successfully decoded the expression, revealing its hidden structure and completing its sequence.

Conclusion

In this mathematical exploration, we've journeyed through the intricacies of binomial expansions, unraveling the pattern within a seemingly complex expression. By applying the binomial theorem and carefully analyzing the given terms, we've successfully identified the missing terms, completing the expression and solidifying our understanding of these fundamental concepts.

This exercise highlights the power of pattern recognition in mathematics. By identifying the underlying structure and applying the appropriate tools, we can solve seemingly intractable problems and gain a deeper appreciation for the elegance and interconnectedness of mathematical principles. The completed expression stands as a testament to our analytical skills and our ability to navigate the world of mathematical puzzles.