Problem 41 - Precalculus with Limits

Question

In Exercises 25 - 30, prove the inequality for the indicated integer values of \( n \). A factor of \( \left(2^{4n - 2} + 1\right) \) is \( 5 \).

Step-by-Step Solution

Answer

Yes, it is proven that a factor of \( \left(2^{4n - 2} + 1\right) \) is \( 5 \). The proof relies on the properties of exponents, the Binominal Theorem and the divisibility rule of 5.

Step 1: Simplify the Equation

Split the exponential term into two parts using the properties of exponents: \( 2^{4n - 2} = (2^{n} - 1) \times (2^{3n} + 2^{2n} + 2^{n} + 1) \)

Step 2: Apply the Binominal Theorem

Applying the Binominal Theorem, the equation expands into: \( 2^{4n} - 2^{3n} + 2^{2n} - 2^{n} + 1 \)

Step 3: Evaluate Divisibility by 5

Notice that except for the \( + 1 \) term at the end, all the other terms in the equation, when squared, ends with a 4 or a 6. Since a number ends in 5 only if it's last digit is 0 or 5 - this can be concluded by the divisibility rule of 5 - it validates that the given number is divisible by 5. In other words, \( \left(2^{4n - 2} + 1\right) \) is indeed a factor of 5.

Key Concepts

Understanding Exponents

Exponents represent the number of times a base number is multiplied by itself. Imagine you have a number like 2 raised to the power of 4, written as \(2^4\). This means you multiply 2 by itself four times: \(2 \times 2 \times 2 \times 2 = 16\).
When working with exponents, it helps to remember some basic rules:

• The product of powers rule: \(a^m \times a^n = a^{m+n}\)
• The power of a power rule: \((a^m)^n = a^{m \cdot n}\)
• The power of a product rule: \((ab)^n = a^n \cdot b^n\)

In our problem, we use these properties to simplify \(2^{4n-2}\) into a more manageable form. Knowing how to manipulate exponents is a key skill when dealing with complex equations and inequalities.

Introduction to the Binomial Theorem

The Binomial Theorem helps expand expressions that are raised to a power. It's useful when dealing with expressions like \((x+y)^n\). The theorem states: \((x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\)
This means you break down the expression into a sum of terms involving the binomial coefficient, \(\binom{n}{k}\). This coefficient represents combinations or ways you can select \(k\) items from \(n\).
In our exercise, the Binomial Theorem allows us to handle complex exponential terms and systematically expand them. Even though we don't directly apply it to \(2^{4n-2}\), understanding it helps us see how polynomials can be manipulated and expanded logically.

Divisibility Rules Simplified

Divisibility rules help quickly determine if one number divides another without performing the full division.
Specifically, the rule for divisibility by 5 tells us a number is divisible by 5 if its last digit is either 0 or 5.
In this exercise, we're checking if \(2^{4n-2} + 1\) is divisible by 5. By observing the last digit of the expression, we can make quick conclusions about divisibility without full calculations.
Divisibility tests are powerful tools, allowing quick checks in problems involving large numbers.

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into simpler polynomials that multiply together to give the original polynomial. It's a reverse process of expanding a polynomial.
For example, the polynomial \(x^2 - 1\) can be factored into \((x-1)(x+1)\).
In our given problem, an understanding of factorization lets us simplify and manipulate the polynomial components of an expression like \(2^{4n-2}\). Factoring helps identify common factors and make the expressions less complex. It also helps us find factors of the polynomial and understand their relationships, which are crucial in solving inequalities.