Question

Use the Binomial Theorem to find the indicated coefficient or term. The coefficient of \(x^{3}\) in the expansion of \((2 x+1)^{12}\)

Short answer

The coefficient of \(x^3\) in the expansion of \((2x + 1)^{12}\) is 1760.

Step-by-step solution

Understand the problem

You are asked to find the coefficient of the term involving \(x^3\) in the expansion of \((2x + 1)^{12}\) using the Binomial Theorem.

Binomial Theorem

Recall the Binomial Theorem: \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]For the expression \((2x + 1)^{12}\), identify \(a = 2x\), \(b = 1\), and \(n = 12\).

Identify the term

We need to find the term where the exponent of \(x\) is 3 in the expansion. Using the general term formula from Binomial Theorem, the \(k\)-th term is given by:\[\binom{12}{k} (2x)^{12-k} (1)^k\]Set the exponent of \(x\) to 3: \[(2x)^{12-k} = x^3 \ 12 - k = 3 \ k = 9\]

Calculate the coefficient

Now substitute \(k = 9\) into the general term formula:\[\binom{12}{9} (2x)^{3} (1)^9 = \binom{12}{9} (2x)^3 = \binom{12}{9} 2^3 x^3\]Simplify the calculations:\[\binom{12}{9} = \binom{12}{3} = \frac{12!}{3! (12-3)!} = \frac{12!}{3! 9!} = 220 \ 2^3 = 8\]

Find the coefficient

Finally, multiply the results to find the coefficient of \(x^3\):\[220 \times 8 = 1760\]

Key concepts

coefficients

In algebra, a coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic term. For instance, in the term *5x*, 5 is the coefficient. In the context of the Binomial Theorem, we often seek specific coefficients of particular terms in expanded polynomial expressions. For our example, the coefficient of **x^3** in \[ (2x + 1)^{12} \] is what we want to find. We use the binomial expansion to break down the exponent in a structured way, allowing us to isolate individual terms and their coefficients. Remember, understanding coefficients helps us interpret how each part of a polynomial behaves in relation to its variables.

expansion

Expansion in mathematics refers to expressing a mathematical expression or equation in an extended form. In the context of the Binomial Theorem, expanding \[ (a + b)^n \] means writing it out as a sum of terms where each term represents a combination of *a* and *b* raised to varying powers. For example, expanding \[ (2x + 1)^{12} \] involves writing it as a sum: \[ (2x + 1)^{12} = \binom{12}{0} (2x)^{12} + \binom{12}{1} (2x)^{11} \times 1 + \binom{12}{2} (2x)^{10} \times 1^2 + \text{...} + \binom{12}{12} \times 1^{12} \]. Each term's coefficient involves a binomial coefficient and helps us isolate the target term—in this case, **x^3**. By systematically expanding the polynomial, we simplify identifying and calculating specific terms.

combinatorics

Combinatorics is a branch of mathematics dealing with combinations, permutations, and the counting of objects. It plays a crucial role in the Binomial Theorem, where we use binomial coefficients, denoted by \[ \binom{n}{k} \], to determine the weights of individual terms in an expanded expression. These coefficients are computed as \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]. For our task, we calculated \[ \binom{12}{9} \], which allows us to identify how many ways we can choose 9 elements from 12—a fundamental combinatorial concept. This combinatorial approach helps us break down and simplify the expanded polynomial into manageable terms.

polynomials

Polynomials are algebraic expressions consisting of variables and coefficients combined using addition, subtraction, and multiplication. Each term in a polynomial is a product of a coefficient and a variable raised to a non-negative integer power. The Binomial Theorem lets us handle polynomials by expanding expressions like \[ (a + b)^n \] into a series of terms. In our example, \[ (2x + 1)^{12} \] is a polynomial raised to the exponent 12. By expanding it, we generate a polynomial with terms involving varied powers of **x**. Understanding polynomials is essential for mastering higher concepts in algebra and calculus, making problems simpler and more structured to solve.