Question

Use the Binomial Theorem to prove that $$2^{n}=\sum_{k=0}^{n}(-1)^{k} C(n, k) 3^{n-k}$$ Write out what the identity says for \(n=4\).

Short answer

Binomial Theorem shows \(2^n=\sum_{k=0}^{n}(-1)^{k} C(n, k) 3^{n-k}\) holds; for \(n=4\), it equals 16.

Step-by-step solution

Understand the Binomial Theorem

The Binomial Theorem states that for any non-negative integer \( n \) and real numbers \( x \) and \( y \), the expansion of \((x + y)^n\) is given by:\[(x + y)^n = \sum_{k=0}^{n} C(n, k) x^{n-k} y^k\]where \( C(n, k) \) are the binomial coefficients, calculated as \( \frac{n!}{k!(n-k)!} \).

Apply the Binomial Theorem to \((1 - 3)^n\)

Consider the expression \((1 - 3)^n\). Using the Binomial Theorem, we expand it as:\[(1 - 3)^n = \sum_{k=0}^{n} C(n, k) 1^{n-k} (-3)^k\]which simplifies to:\[(-2)^n = \sum_{k=0}^{n} C(n, k) (-3)^k\]

Simplify the Expression

Recognize that \((-2)^n\) can also be written using powers of -1 and 2:\[(-1)^n 2^n = \sum_{k=0}^{n} C(n, k) (-1)^k 3^k\]This is equivalent to:\[2^n = \sum_{k=0}^{n} (-1)^{k} C(n, k) 3^{n-k}\]by multiplying both sides by \((-1)^n\).

Verify the Identity for n=4

For \( n = 4 \), we compute:\[2^4 = 16\]Now calculate the right-hand side of the identity:\[\sum_{k=0}^{4} (-1)^{k} C(4, k) 3^{4-k}\]= \(3^4 - 4 \times 3^3 + 6 \times 3^2 - 4 \times 3^1 + 3^0\)= \(81 - 108 + 54 - 12 + 1 = 16\).The verified result confirms the identity.

Key concepts

Binomial Coefficients

In combinatorics, binomial coefficients arise in the expansion of a binomial raised to a power. When you see an expression like \( (x + y)^n \), the binomial coefficients \( C(n, k) \) tell us how many ways we can choose \( k \) elements from \( n \) elements. These coefficients are key to the Binomial Theorem and are calculated as:

• \( C(n, k) = \frac{n!}{k!(n-k)!} \)

Here, \( n! \) (n factorial) refers to the product of all positive integers up to \( n \). This coefficient appears in many mathematical fields, especially when dealing with expansions and counting problems. Understanding how to compute and apply them is crucial when using the Binomial Theorem.

Expansion

Expansion in the context of the Binomial Theorem refers to expressing a binomial \((x + y)^n\) as a sum of terms involving binomial coefficients. This expansion is outlined by:

• \( (x + y)^n = \sum_{k=0}^{n} C(n, k) x^{n-k} y^k \)

The idea is to transform a simple expression into a greater, detailed form. For instance, consider \( (1 - 3)^n \). Using the theorem gives us a way to expand it thoroughly:

• \( (1 - 3)^n = \sum_{k=0}^{n} C(n, k) 1^{n-k} (-3)^k \)

This simplification allows us to manage complex expressions by breaking them into manageable pieces.

Proof by Induction

Proof by induction is a logical approach used to prove mathematical statements, often involving a variable integer \( n \). The process involves two critical steps:

• Base Case: Verify the statement for the initial value, often \( n = 0 \) or \( n = 1 \). This serves as the foundation for the entire proof.
• Inductive Step: Assume the statement holds for an arbitrary positive integer \( n = k \). Then, prove it holds for \( n = k+1 \) based on this assumption.

This technique is essential for proving identities like the given binomial expression. Each step relies on the truth from the prior, effectively creating a domino effect confirming the statement's validity for all integers \( n \). In our case, the setup through the Binomial Theorem and verifying for \( n = 4 \) is part of reinforcing this identity methodically. Proof by induction provides a powerful framework for establishing truths in mathematics, especially involving infinite sequences or series.