Question

Find the independent term in the expansion of \(\left(x^{4}+\frac{3}{8 x^{3} \sqrt{x}}\right)^{15}\) (1) \({ }^{15} \mathrm{C}_{4}\left(\frac{3}{8}\right)^{16}\) (2) \({ }^{15} \mathrm{C}_{12}\left(\frac{3}{8}\right)^{4}\) (3) \({ }^{15} \mathrm{C}_{8}\left(\frac{3}{8}\right)^{8}\) (4) \({ }^{15} \mathrm{C}_{7}\left(\frac{3}{8}\right)^{12}\)

Short answer

Answer: There is no independent term in the expansion.

Step-by-step solution

Simplify the second term

Before applying the binomial theorem, let's simplify the second term: \(b = \frac{3}{8 x^{3}\sqrt{x}} = \frac{3}{8 x^{7/2}} = \frac{3}{8}x^{-7/2}\)

Apply the binomial theorem

Now, we apply the binomial theorem: \((a+b)^{15} = \sum_{k=0}^{15}\binom{15}{k}a^{15-k}b^k\)

Find the term with no x component

To find the independent term, we need to look for the term with \(x^0\). Let's consider a generic term in the expansion: \(\binom{15}{k}(x^4)^{15-k}\left(\frac{3}{8}x^{-7/2}\right)^k\). The power of x in this term is \(4(15-k)-\frac{7}{2}k\). In order to get a term with no x component, we need to find k such that: \(4(15-k)-\frac{7}{2}k = 0\) Solving this equation for k, we get: \(k = \frac{60}{11}\) However, k must be an integer, and therefore, there is no independent term in the expansion. In this case, none of the provided answers are correct since there is no independent term in the expansion.

Key concepts

Independent Term

Understanding the concept of an independent term in a binomial expansion is crucial for solving various algebraic problems. The independent term is essentially an element within the expanded form of a binomial that does not contain the variable being considered.

Let's consider the binomial expansion \( (a + b)^n \) where \( a \) and \( b \) are terms that may include a variable, say \( x \), and \( n \) is a non-negative integer. Upon expanding the binomial using the Binomial Theorem, the expansion will consist of several terms. Each of these terms will have \( x \) raised to some power, except possibly one term where the power of \( x \) is zero, making it the independent term.

To identify the independent term, one must equate the overall exponent of \( x \) in any term to zero and solve for the binomial coefficient or index. This process involves finding a specific \( k \) such that \( x \) disappears or its exponent becomes zero. However, if no suitable integer satisfies this condition, as in the example exercise, the conclusion is that there is no independent term in the given binomial expansion.

Binomial Expansion

Getting to grips with binomial expansion is a foundational skill in algebra that unlocks the understanding of polynomial expressions. The Binomial Theorem provides a formula for expanding expressions that are raised to a power and are in the form of \( (a + b)^n \).

In the formula \( \binom{n}{k} a^{n-k} b^k \), \( \binom{n}{k} \) denotes a binomial coefficient and represents the number of ways to choose \( k \) elements from a set of \( n \) elements without considering the order. This coefficient is also represented as \( ^nC_k \). The expansion is a sum of terms, where the exponents of \( a \) and \( b \) in each term are complementary; they add up to \( n \).

For example, in the binomial expansion of \( (a+b)^3 \) the expanded form is: \( a^3 + 3a^2b + 3ab^2 + b^3 \). It follows a pattern where the exponent of \( a \) decreases and the exponent of \( b \) increases in each successive term. This pattern is systematic and predictable, which allows one to write out the terms of a binomial expansion without necessarily multiplying everything out.

Mathematical Induction

Mathematical induction is a powerful proof technique used to establish the truth of an infinite number of statements. The process involves two primary steps: the base case and the induction step.

The base case confirms that the statement holds true for the initial value, usually \( n = 1 \). In the induction step, the assumption is that the statement holds for some arbitrary positive integer \( n \), and then it's shown that the statement is true for \( n + 1 \) as well.

This technique is akin to a domino effect; knocking over the first domino (the base case) and showing that one domino falling will cause the next one to fall (the induction step) proves that all the dominoes will fall. In mathematics, this is used to prove propositions, theorems, or formulas about integers, often involving expressions or sequences. For example, the principle of mathematical induction can demonstrate the validity of the binomial theorem itself for all non-negative integer exponents. It serves as a fundamental tool for establishing the truth of statements where direct verification isn't feasible due to the infinite nature of the set being considered.