Binomial Theorem Expansion Problems And Solutions

The realm of mathematics is filled with fascinating concepts, and one of the most powerful tools in our arsenal is the binomial theorem. This theorem provides a systematic way to expand expressions of the form $(x + y)^n$, where $n$ is a non-negative integer. In this article, we will delve into the intricacies of binomial expansion by tackling two intriguing problems. These problems will not only test your understanding of the binomial theorem but also showcase its versatility in solving various mathematical challenges. Let's embark on this journey of mathematical exploration and unlock the secrets of binomial expansion.

Problem 1: Deciphering the Coefficients in a Binomial Product Expansion

Understanding the Problem Statement

Our first challenge involves an expression that combines a linear term and a binomial raised to a power: $(2-x)(1+ax)^6$. The problem states that when this expression is expanded in ascending powers of $x$, up to the term in $x^2$, it takes the form $1 + bx^2 + \cdots$. Our mission is to find the values of the constants $a$ and $b$. This problem beautifully illustrates the interplay between binomial expansion and algebraic manipulation. To solve it, we will need to carefully expand the binomial term, multiply it with the linear term, and then compare the coefficients of the resulting expression with the given form. This process will lead us to a system of equations that we can solve to find the values of $a$ and $b$.

Applying the Binomial Theorem

To tackle this problem effectively, we must first recall the binomial theorem. The binomial theorem states that for any non-negative integer $n$ and any real numbers $x$ and $y$:

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$

where $\binom{n}{k}$ represents the binomial coefficient, also known as the "n choose k" coefficient, and is calculated as:

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

In our case, we need to expand $(1 + ax)^6$. Applying the binomial theorem, we get:

$(1 + ax)^6 = \binom{6}{0}1^6(ax)^0 + \binom{6}{1}1^5(ax)^1 + \binom{6}{2}1^4(ax)^2 + \cdots$

Let's calculate the first few binomial coefficients:

• $\binom{6}{0} = \frac{6!}{0!6!} = 1$
• $\binom{6}{1} = \frac{6!}{1!5!} = 6$
• $\binom{6}{2} = \frac{6!}{2!4!} = \frac{6 \times 5}{2} = 15$

Substituting these values back into the expansion, we get:

$(1 + ax)^6 = 1 + 6ax + 15a^2x^2 + \cdots$

Multiplying and Collecting Terms

Now that we have expanded the binomial term, we need to multiply it by $(2 - x)$:

$(2 - x)(1 + ax)^6 = (2 - x)(1 + 6ax + 15a^2x^2 + \cdots)$

Expanding this product, we get:

$2(1 + 6ax + 15a^2x^2 + \cdots) - x(1 + 6ax + 15a^2x^2 + \cdots)$

$= 2 + 12ax + 30a^2x^2 + \cdots - x - 6ax^2 - 15a^2x^3 - \cdots$

Now, let's collect the terms with the same powers of $x$:

$= 2 + (12a - 1)x + (30a^2 - 6a)x^2 + \cdots$

Equating Coefficients and Solving for a and b

The problem states that the expansion is of the form $1 + bx^2 + \cdots$. Comparing this with our expanded form, we can equate the coefficients:

• Constant term: $2 = 1$ (This is a contradiction, indicating a potential issue with the problem statement or a misunderstanding. We will address this later.)
• Coefficient of $x$: $12a - 1 = 0$
• Coefficient of $x^2$: $30a^2 - 6a = b$

From the equation $12a - 1 = 0$, we can solve for $a$:

$12a = 1$

$a = \frac{1}{12}$

Now that we have the value of $a$, we can substitute it into the equation for $b$:

$b = 30(\frac{1}{12})^2 - 6(\frac{1}{12})$

$b = 30(\frac{1}{144}) - \frac{6}{12}$

$b = \frac{30}{144} - \frac{1}{2}$

$b = \frac{5}{24} - \frac{12}{24}$

$b = -\frac{7}{24}$

Addressing the Contradiction and Refining the Solution

We encountered a contradiction when equating the constant terms: $2 = 1$. This indicates that there might be a typo in the problem statement. It's highly likely that the expanded form should have a constant term of 2, matching the constant term in our expansion. Assuming the correct form is $2 + bx^2 + \cdots$, our solution for $a$ and $b$ remains valid.

Therefore, the values are:

$a = \frac{1}{12}$

$b = -\frac{7}{24}$

Conclusion for Problem 1

In this problem, we successfully navigated the intricacies of binomial expansion and algebraic manipulation. We meticulously expanded the given expression, equated coefficients, and solved for the unknown constants. The importance of careful calculations and attention to detail cannot be overstated in such problems. Despite encountering a potential issue with the problem statement, we were able to refine our approach and arrive at a logical solution. This problem serves as a testament to the power of the binomial theorem and its applications in various mathematical contexts.

Problem 2: Unveiling the Initial Terms of a Binomial Expansion

Deciphering the Problem Statement

Our second challenge shifts our focus to finding the first three terms in the expansion of $(1 + a)^{10}$ in ascending powers of $a$. This problem is a direct application of the binomial theorem, allowing us to showcase its elegance and efficiency. We will systematically apply the theorem, calculate the relevant binomial coefficients, and construct the desired terms. This exercise will further solidify our understanding of binomial expansion and its ability to provide concise representations of complex expressions.

Applying the Binomial Theorem

As we did in Problem 1, we will again rely on the binomial theorem to solve this problem. The theorem states that for any non-negative integer $n$ and any real numbers $x$ and $y$:

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$

In this case, we want to expand $(1 + a)^{10}$. Applying the binomial theorem, we get:

$(1 + a)^{10} = \binom{10}{0}1^{10}a^0 + \binom{10}{1}1^9a^1 + \binom{10}{2}1^8a^2 + \cdots$

We only need to find the first three terms, so we will focus on the terms corresponding to $k = 0, 1,$ and $2$. Let's calculate the required binomial coefficients:

• $\binom{10}{0} = \frac{10!}{0!10!} = 1$
• $\binom{10}{1} = \frac{10!}{1!9!} = 10$
• $\binom{10}{2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2} = 45$

Constructing the First Three Terms

Now that we have the binomial coefficients, we can substitute them back into the expansion:

$(1 + a)^{10} = 1 \cdot 1^{10} \cdot a^0 + 10 \cdot 1^9 \cdot a^1 + 45 \cdot 1^8 \cdot a^2 + \cdots$

Simplifying, we get:

$(1 + a)^{10} = 1 + 10a + 45a^2 + \cdots$

Therefore, the first three terms in the expansion of $(1 + a)^{10}$ in ascending powers of $a$ are $1$, $10a$, and $45a^2$.

Conclusion for Problem 2

This problem provided a straightforward yet elegant application of the binomial theorem. By systematically calculating the binomial coefficients and substituting them into the expansion, we efficiently determined the first three terms of the expansion. This exercise reinforces the power and elegance of the binomial theorem in simplifying complex expressions. It also highlights the importance of understanding and applying mathematical tools to solve problems effectively.

In this article, we have explored the fascinating world of binomial expansion through two challenging problems. We have seen how the binomial theorem provides a powerful tool for expanding expressions of the form $(x + y)^n$. By meticulously applying the theorem, calculating binomial coefficients, and performing algebraic manipulations, we were able to solve complex problems and gain deeper insights into the properties of binomial expansions. These problems underscore the importance of understanding fundamental mathematical principles and their applications in various contexts. As we continue our journey in mathematics, let us remember the power of binomial expansion and its ability to unlock new mathematical horizons.