Question

Find the independent term in the expansion of \(\left(5 x^{2}-\frac{1}{x^{4}}\right)^{6}\). (1) 8250 (2) 8560 (3) 9250 (4) 9375

Short answer

Question: Find the independent term in the expansion of \((5x^2 - \frac{1}{x^4})^6\). Answer: 9375

Step-by-step solution

Understand Binomial Theorem

The binomial theorem states that for any positive integer \(n\), and any real numbers \(a\) and \(b\), we have: \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k\) Where \(\binom{n}{k}\) is the binomial coefficient, which can be computed as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

Identify a and b

From the given expression, we have \(a = 5x^2\) and \(b = -\frac{1}{x^4}\).

Find the term with no variable x

We need to find the exponent \(k\) for which the term \(a^{n-k}b^k\) has no variable \(x\). We have, \(a^{n-k}b^k = (5x^2)^{6-k} \left(-\frac{1}{x^4}\right)^k\). Multiplying the exponents, we get: \(a^{n-k} b^k = 5^{6-k} x^{12-2k} (-1)^k x^{-4k}\). Combining the exponents of \(x\), \(a^{n-k} b^k = 5^{6-k} (-1)^k x^{12-6k}\) Now, we need to find the value of k such that the exponent of x is zero, i.e., \(12 - 6k = 0\)

Solve for k

Solving the equation \(12 - 6k = 0\) for k, we get: \(6k = 12\) \(k = \frac{12}{6}\) \(k = 2\)

Compute the independent term

Now that we have the value of \(k\), we can find the independent term using the binomial theorem formula: Independent term \(= \binom{n}{k} a^{n-k}b^k\) \(= \binom{6}{2} (5x^2)^{6-2} \left(-\frac{1}{x^4}\right)^2\) \(= \frac{6!}{2!(6-2)!} (5^4) (1)^2\) \(= \frac{6!}{2!4!} (625)\) \(= \frac{720}{48} (625)\) \(= 15(625)\) \(= 9375\) The independent term in the expansion is 9375, which corresponds to option (4).

Key concepts

Independent Term

An independent term in a mathematical expansion is a term free of any variables. In polynomial expressions, especially those derived from expansions using the Binomial Theorem, the independent term is the constant term. This specific term does not have any variable attached, meaning the exponent of the variable is zero.

In the context of finding the independent term in an expansion, such as \( \left(5x^{2}-\frac{1}{x^{4}}\right)^{6} \), the process involves finding the specific combination of terms where all variables cancel out. This signifies that the exponent of the variable is zero.

For instance, when searching for the independent term, you solve for a particular power of the term in the binomial expansion that results in the final expression with no variables present, like setting up the equation \( 12 - 6k = 0 \) to find the correct value of \( k \). This step is crucial to isolate the numeric 'constant' from the variable expression.

Binomial Coefficient

The binomial coefficient is a fundamental element of binomial expansion, denoted as \( \binom{n}{k} \), which represents the number of ways to choose \( k \) elements from a set of \( n \) elements without considering the order. It is a key player in the expression of a binomial expansion using the formula: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k \).

This coefficient is calculated using factorials: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
For example, when calculating the independent term of \( \left(5x^{2}-\frac{1}{x^{4}}\right)^{6} \), and needing \( k=2 \), you compute \( \binom{6}{2} \). This involves finding the factorial of 6, then dividing it by the factorial of 2 and the factorial of 4 (since \( 6-2 = 4 \)).

The role of binomial coefficients is crucial as it scales each term's contribution to the final expansion, essentially dictating the number of ways the particular term can appear in the expansion.

Mathematical Expansion

Mathematical expansions involve expressing a complex expression in simpler ones. Specifically, in the context of the Binomial Theorem, a binomial expression like \( (a + b)^n \) is expanded into a sum of terms calculated using the formula: \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k \).

This expansion breaks down the complex expression into multiple components, each accompanied by a binomial coefficient, and raised to powers that summarize the possibilities of choosing terms with certain powers.

In our exercise, expanding \( \left(5x^{2}-\frac{1}{x^{4}}\right)^{6} \) yields various terms combining powers of \(5x^{2} \) and \(-\frac{1}{x^{4}} \). Each term in this expansion is systematically derived and calculated using particular values for \( k \) (the term number), and the entirety of the sum is then considered, isolating any term needed (in this case, the independent term).

Such expansions are widely used to simplify polynomials and to find specific terms without necessarily listing all expanded terms.

Exponent Rules

Exponent rules are critical in simplifying expressions during expansions, particularly when dealing with terms like \( (5x^{2})^{6-k} \left(-\frac{1}{x^{4}}\right)^k \). These rules help in combining and simplifying powers of base terms.

Some essential exponent rules include:

• Multiplying exponents when raising a power to another power: \( (a^m)^n = a^{m\cdot n} \).
• Dividing coefficients when simplifying terms: \( \frac{1}{x^n} = x^{-n} \).
• Combining exponents when multiplying like bases: \( x^a \cdot x^b = x^{a+b} \).

For instance, when handling terms like \( (5x^2)^{6-k} \) and \( \left(-\frac{1}{x^4}\right)^k \), you apply these rules to simplify and eventually find the expression of \( x^{12-6k} \).

Understanding these rules assists in finding specific terms and, in our case, the independent term by setting equations like \( 12 - 6k = 0 \) to zero, signifying no variable factor remains.