Problem 28 - Discrete Mathematics with Applications

Question

Using the binomial theorem, prove each. \(2^{4 n}+3 n-1\) is divisible by \(9 .\) (Hint: \(2=3-1 .\))

Step-by-Step Solution

Answer

Using the binomial theorem, we expand and simplify the expression \((3-1)^{4n}\). We then separate the terms for which k is even and k is odd, and remove even factors from the terms with even k. By rearranging the expression and adding \(3n\) and \(-1\) on both sides of the equation, we demonstrate that the given expression is divisible by 9 as the sum of two expressions that are both divisible by 9, as well as the terms \(3n\) and \(-1\). Therefore, \(2^{4 n}+3 n-1\) is divisible by 9.

Step 1: Apply the binomial theorem to the expression

Using the binomial theorem, we expand the expression \((3-1)^{4n}\): \((3-1)^{4n} = \sum_{k=0}^{4n} \binom{4n}{k} 3^{4n-k} (-1)^k\)

Step 2: Simplify the expression

As \(3-1=2\), the given expression becomes: \(2^{4n} = \sum_{k=0}^{4n} \binom{4n}{k} 3^{4n-k} (-1)^k\)

Step 3: Separate the terms for which k is even and k is odd

We can write two separate sums for the terms in the expansion with \(k\) being even and \(k\) being odd: \(2^{4n} = \sum_{k=0}^{2n} \binom{4n}{2k} 3^{4n-2k} (-1)^{2k} - \sum_{k=0}^{2n-1} \binom{4n}{2k+1} 3^{4n-2k-1} (-1)^{2k+1}\)

Step 4: Remove even factors from the terms with even k

Since all the even factors in the expansion have exponent \((-1)^{2k}\), which equals \(1\) for each term, we can remove them without changing the value of the expression: \(2^{4n} = \sum_{k=0}^{2n} \binom{4n}{2k} 3^{4n-2k} - \sum_{k=0}^{2n-1} \binom{4n}{2k+1} 3^{4n-2k-1} (-1)^{2k+1}\)

Step 5: Rearrange the expression and add 3n and -1 on both sides of the equation

Now rearrange the formula to get: \(2^{4 n}+3 n-1 = \left(\sum_{k=0}^{2n} \binom{4n}{2k} 3^{4n-2k}\right) - 3n +3n-1 + \left(-\sum_{k=0}^{2n-1} \binom{4n}{2k+1} 3^{4n-2k-1} (-1)^{2k+1}\right)\)

Step 6: Demonstrate that the given expression is divisible by 9

Now, since each term in the first sum is divisible by 3, the difference between the consecutive terms in the first sum is divisible by \(3^2=9\). Therefore, the whole first sum must be divisible by 9. Similarly, each term in the second sum is divisible by 3, and the difference between consecutive terms is divisible by \(3^2=9\). Therefore, the whole second sum must also be divisible by 9. The given expression is the sum of two expressions that are both divisible by 9, as well as the terms \(3n\) and \(-1\) which are also divisible by 9, therefore the whole expression \(2^{4 n}+3 n-1\) is divisible by 9.

Key Concepts

Divisibility Proof

The concept of proving divisibility is crucial for verifying that one number can exactly divide another without leaving a remainder. In this exercise, we are asked to prove that the expression \(2^{4n} + 3n - 1\) is divisible by 9.

Divisibility proofs usually involve several logical steps and can be achieved using different methods. One effective method is the direct approach, which examines the factors and properties of numbers involved. In our problem, we expanded \((3-1)^{4n}\) using the binomial theorem and showed that both the sums resulting from the expansion lead to components divisible by 9.

Understanding divisibility in mathematical expressions ensures that complex equations can be simplified by considering their divisible properties, often by a constant like 9 in this case.

Mathematical Induction

Mathematical induction is a powerful technique often used to prove the validity of statements involving integers, particularly when they involve a sequence or pattern. Though this problem was solved using the binomial theorem, mathematical induction could also serve as a method to affirm the statement.

The inductive reasoning process consists of two main steps:

• Base Case: Verify the statement is true for the initial value, typically \(n=1\).
• Inductive Step: Assume the statement is true for \(n=k\), then prove it holds for \(n=k+1\).

If both steps are successfully shown, it implies the statement is valid for all integers. While induction was not directly applied in the solution, it complements the notion of systematically proving divisibility over integers.

Combinatorial Identities

Combinatorial identities involve expressions that equate different mathematical combinations or sums. They often appear in problems relating to permutations, combinations, and binomial expansions.

In this exercise, we leveraged a combinatorial identity from the binomial theorem to transform the power of a binomial \((3-1)^{4n}\). This expansion used coefficients \(\binom{4n}{k}\) reflective of the number of ways to choose \(k\) items from \(4n\) options, which is inherent in binomial expansions.

Recognizing and using these identities allows for the simplification of complex algebraic expressions and sometimes directly leads to proofs via properties involving binomial coefficients. Understanding these fundamentals helps in solving problems efficiently and elegantly, proving not just divisibility but also equating expressions in multifaceted scenarios.