Rational Terms In Binomial Expansion Of (4^(1/4) + 5^(1/6))^120

In the realm of mathematics, particularly within the study of binomial expansions, a fascinating problem arises when considering the number of rational terms present in the expansion of expressions involving radicals. This article delves into a specific instance of such a problem, focusing on the binomial expansion of ${ (4^{\frac{1}{4}} + 5^{\frac{1}{6}})^{120} }$. Our goal is to meticulously determine the count of terms that yield rational numbers upon expansion. This exploration necessitates a firm understanding of the binomial theorem, properties of exponents and radicals, and the conditions under which terms in an expansion become rational. The journey through this problem will not only illuminate the solution but also reinforce fundamental concepts in algebra and number theory. Let's embark on this mathematical endeavor, unraveling the intricacies of binomial expansions and rational terms. This exploration is crucial for anyone studying advanced algebra, number theory, or preparing for mathematical competitions. Understanding the nature of binomial expansions and how to identify rational terms is a cornerstone in many areas of mathematics and its applications. We will break down the problem step by step, providing a clear and comprehensive solution along with the reasoning behind each step. This article aims to make the solution accessible to students and enthusiasts alike, fostering a deeper appreciation for the elegance and power of mathematical techniques.

The binomial theorem provides a powerful tool for expanding expressions of the form ${(a + b)^n}$, where ${n}$ is a non-negative integer. The theorem states that:

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

Where ${\binom{n}{k}}$ represents the binomial coefficient, calculated as:

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

This coefficient gives the number of ways to choose ${k}$ items from a set of ${n}$ items. The binomial theorem essentially breaks down the expansion of ${(a + b)^n}$ into a sum of terms, each consisting of a binomial coefficient, a power of ${a}$, and a power of ${b}$. For our specific problem involving ${ (4^{\frac{1}{4}} + 5^{\frac{1}{6}})^{120} }$, understanding the binomial theorem is the first step in determining the number of rational terms. Applying the theorem allows us to express the given expression as a sum of individual terms, each of which we can then analyze for rationality. This theorem is not just a formula; it is a fundamental concept in algebra with wide-ranging applications, from probability theory to calculus. The beauty of the binomial theorem lies in its ability to systematically expand complex expressions, making them easier to analyze and manipulate. To fully grasp the solution to our problem, it's imperative to have a solid foundation in the binomial theorem and its implications. This understanding will enable us to navigate the complexities of exponents, radicals, and rational numbers within the context of the expansion.

To determine the number of rational terms in the expansion of ${ (4^{\frac{1}{4}} + 5^{\frac{1}{6}})^{120} }$, we first apply the binomial theorem. Let ${ a = 4^{\frac{1}{4}} }$ and ${ b = 5^{\frac{1}{6}} }$, and ${ n = 120 }$. Using the binomial theorem, we have:

${ (4^{\frac{1}{4}} + 5^{\frac{1}{6}})^{120} = \sum_{k=0}^{120} \binom{120}{k} (4^{\frac{1}{4}})^{120-k} (5^{\frac{1}{6}})^k }$

Now, let's simplify the terms inside the summation:

${ (4^{\frac{1}{4}})^{120-k} = 4^{\frac{120-k}{4}} = (2^2)^{\frac{120-k}{4}} = 2^{\frac{120-k}{2}} }$

${ (5^{\frac{1}{6}})^k = 5^{\frac{k}{6}} }$

Thus, the general term in the expansion can be written as:

${ T_{k+1} = \binom{120}{k} 2^{\frac{120-k}{2}} 5^{\frac{k}{6}} }$

This expression represents the ${(k+1)}$-th term in the expansion. To find the rational terms, we need to identify the values of ${k}$ for which the exponents of both 2 and 5 are integers. This step is crucial because a number is rational if and only if it can be expressed as a fraction where both the numerator and denominator are integers. In our case, the exponents determine the rationality of the terms. The binomial coefficient ${\binom{120}{k}}$ is always an integer, so we only need to focus on the exponents of 2 and 5. Understanding how the binomial theorem transforms the original expression into a sum of terms is the key to solving this problem. By isolating the general term, we can focus on the conditions that make it rational, which involves the exponents of the radical terms.

For a term in the expansion to be rational, the exponents of both 2 and 5 in the general term

${ T_{k+1} = \binom{120}{k} 2^{\frac{120-k}{2}} 5^{\frac{k}{6}} }$

must be integers. This means that ${\frac{120-k}{2}}$ and ${\frac{k}{6}}$ must be integers. Let's analyze these conditions separately.

Condition 1: ${\frac{120-k}{2}}$ is an integer

For ${\frac{120-k}{2}}$ to be an integer, ${120-k}$ must be divisible by 2. Since 120 is an even number, ${k}$ must also be an even number. This can be expressed as:

${ k = 2m }$

where ${m}$ is an integer.

Condition 2: ${\frac{k}{6}}$ is an integer

For ${\frac{k}{6}}$ to be an integer, ${k}$ must be divisible by 6. This can be expressed as:

${ k = 6n }$

where ${n}$ is an integer.

Combining both conditions, we need to find values of ${k}$ that satisfy both ${k = 2m}$ and ${k = 6n}$. This implies that ${k}$ must be a multiple of both 2 and 6. The least common multiple (LCM) of 2 and 6 is 6. Therefore, ${k}$ must be a multiple of 6. This condition significantly narrows down the possible values of ${k}$. By focusing on the conditions required for rationality, we transform the problem into a number theory question, specifically concerning divisibility and multiples. The interplay between the exponents and the divisibility rules is crucial for identifying the rational terms in the binomial expansion. This analysis provides a clear pathway to determining the valid values of ${k}$ and, consequently, the number of rational terms.

From the previous section, we established that for a term to be rational, ${k}$ must be a multiple of 6. Since ${0 \leq k \leq 120}$, we can express ${k}$ as:

${ k = 6n }$

where ${n}$ is an integer. Now, we need to find the possible values of ${n}$ that satisfy the condition ${0 \leq k \leq 120}$. Substituting ${k = 6n}$, we get:

${ 0 \leq 6n \leq 120 }$

Dividing all sides by 6, we have:

${ 0 \leq n \leq 20 }$

This inequality tells us that ${n}$ can take integer values from 0 to 20, inclusive. Each value of ${n}$ corresponds to a value of ${k}$ that makes the term in the binomial expansion rational. Therefore, the possible values of ${n}$ determine the number of rational terms. The range of ${n}$ from 0 to 20 gives us a total of 21 possible values (0, 1, 2, ..., 20). This step is pivotal in bridging the gap between the divisibility conditions and the actual count of rational terms. By translating the constraint on ${k}$ into a constraint on ${n}$, we can easily enumerate the possible values and arrive at the final answer. The systematic approach of setting up the inequality and solving for ${n}$ ensures that we capture all valid values of ${k}$ and none that would lead to irrational terms.

Since ${n}$ can take 21 integer values (from 0 to 20), there are 21 values of ${k}$ that make the terms in the binomial expansion of ${ (4^{\frac{1}{4}} + 5^{\frac{1}{6}})^{120} }$ rational. Each value of ${k}$ corresponds to a rational term in the expansion. Therefore, the number of rational terms in the binomial expansion is 21.

In conclusion, by applying the binomial theorem and analyzing the conditions for rationality, we have determined that there are 21 rational terms in the binomial expansion of ${ (4^{\frac{1}{4}} + 5^{\frac{1}{6}})^{120} }$. This problem highlights the interplay between algebraic expansions and number theory concepts. The key to solving this problem lies in understanding the binomial theorem, identifying the conditions for terms to be rational, and systematically counting the values of ${k}$ that satisfy those conditions. The process involved breaking down the problem into manageable steps, starting with the application of the binomial theorem and progressing through the simplification of terms, the identification of divisibility conditions, and the final enumeration of solutions. This approach not only provides the answer but also reinforces a valuable problem-solving strategy applicable to various mathematical challenges. The determination of rational terms in binomial expansions is a classic problem that showcases the elegance and interconnectedness of mathematical ideas. By mastering the techniques demonstrated in this article, students and enthusiasts can confidently tackle similar problems and deepen their understanding of algebra and number theory. The final answer, 21 rational terms, is a testament to the power of mathematical reasoning and the beauty of binomial expansions.