Question

Find the coefficient of \(\mathrm{x}^{7}\) in the expansion of \(\left(7 \mathrm{x}+\frac{2}{\mathrm{x}^{2}}\right)^{13}\). (1) \(78 \times 8^{*} \times 4\) (2) \(78 \times 7^{6} \times 4^{2}\) (3) \(78 \times 7^{11} \times 4\) (4) \(78 \times 7^{11} \times 4^{2}\)

Short answer

Answer: The coefficient of \(x^7\) in the expansion is \(78 \times 7^{11} \times 4\).

Step-by-step solution

Identify the term containing x^7

In order to find the term containing \(x^7\), we will use the binomial theorem, which states that \((x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k\). In this case, \(x = 7x\) and \(y = \frac{2}{x^2}\). We need to find a k such that \((7x)^{13-k}\left(\frac{2}{x^2}\right)^k\) contains \(x^7\).

Determine the value of k

We need to find the value of k such that the term \((7x)^{13-k}\left(\frac{2}{x^2}\right)^k\) has a power of 7 in x. The power of x in this term is \(13 - k - 2k = 7\). Solving for k, we get \(13 - 3k = 7\), so \(k = 2\).

Evaluate the term with the given k value

Having found the value of k that yields a term with an x power of 7, we can now evaluate the term with k = 2. Using the binomial theorem, the term is \(\binom{13}{2}(7x)^{13-2}\left(\frac{2}{x^2}\right)^2 = \binom{13}{2} (7x)^{11} \cdot \frac{4}{x^4}\).

Calculate the coefficient

Now that we have found the term including \(x^7\), we can calculate its coefficient. The coefficient is given by \(\binom{13}{2} \cdot 7^{11} \cdot \frac{4}{x^4}\) after simplifying our term obtained in the previous step. Evaluating this expression, we get \(\binom{13}{2} \cdot 7^{11} \cdot 4\). Since \(\binom{13}{2} = 78\), the coefficient is \(78 \cdot 7^{11} \cdot 4\). So, the correct answer is (3) \(78 \times 7^{11} \times 4\).

Key concepts

Understanding Binomial Expansion

The binomial expansion is a powerful algebraic tool used to expand expressions that are raised to a power. When you have a binomial like \((a + b)^n\), the binomial theorem provides a way to expand it into a series:
\[(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\]
Here, \(n\) is the exponent, and \(\binom{n}{k}\) is the binomial coefficient, which determines how many ways you can choose \(k\) elements from a set of \(n\) elements. This allows us to break down complex expressions into simpler terms of constant and variable powers.

• \(a^{n-k}\) and \(b^k\) are the terms with decreasing and increasing powers respectively.
• The sum of these powers always equals \(n\).

In the exercise, the expression \((7x + \frac{2}{x^2})^{13}\) is expanded using this theorem.

Calculating Coefficients in Binomial Expansion

Finding specific coefficients in a binomial expansion is crucial, especially when dealing with polynomial expressions. The binomial coefficient \(\binom{n}{k}\) represents the number of terms, or ways, each term of the expansion can be formed. It is calculated using the formula:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
Where \(n!\) (n factorial) is the product of all positive integers up to \(n\). In our example, we need to find the term containing \(x^7\).

• The term \((7x)^{13-k} \left(\frac{2}{x^2}\right)^k\) must be evaluated to isolate \(x^7\).
• Setting the powers of \(x\) in this term equal to 7 helps identify \(k\): \(13 - 3k = 7\), leading to \(k = 2\).

Once \(k\) is known, calculate \(\binom{13}{2}\) to find the coefficient, and multiply with other constant terms as per the binomial expression structure.

Identifying the Power of a Term

When expanding binomials, identifying the power of a specific term, like \(x^7\), involves careful algebraic manipulation. You must determine how the expression's variables and exponents contribute to the desired power.

• The task is to balance terms to achieve the required variable power by solving the equation derived from the exponents.
• In the expression \((7x)^{13-k}\left(\frac{2}{x^2}\right)^k\), we identified that the power of \(x\) is \(13 - k - 2k\).

Using the equation \(13 - 3k = 7\), we solved for \(k = 2\). This allows us to substitute it back into the binomial terms to calculate the desired power. This highlights how each factor in the expansion influences the power of \(x\) within the broader polynomial structure.