Problem 14 - College Algebra with Modeling and Visualization

Question

Use mathematical induction to prove the statement. Assume that \(n\) is a positive integer. $$ x^{2 n}+x^{2 n-1} y+\cdots+x y^{2 n-1}+y^{2 n}=\frac{x^{2 n+1}-y^{2 n+1}}{x-y} $$

Step-by-Step Solution

Answer

The statement is true by mathematical induction for all positive integers \( n \).

Step 1: Base Case

For mathematical induction, we start by verifying the base case where \( n = 1 \). Substitute \( n = 1 \) into the given equation:\[x^2 + xy + y^2 = \frac{x^{3} - y^{3}}{x - y}\]The right-hand side can be expanded as:\[\frac{(x-y)(x^2+xy+y^2)}{x-y} = x^2 + xy + y^2\]Both sides are equal, thus the base case holds true.

Step 2: Inductive Step – Assume True for n=k

Assume the statement is true for \( n = k \), meaning:\[x^{2k} + x^{2k-1}y + \, \cdots \, + xy^{2k-1} + y^{2k} = \frac{x^{2k+1} - y^{2k+1}}{x-y}\]

Step 3: Inductive Step – Prove for n=k+1

We need to show the statement holds for \( n = k+1 \):The left-hand side becomes:\[x^{2(k+1)} + x^{2(k+1)-1}y + \, \cdots \, + xy^{2(k+1)-1} + y^{2(k+1)}\]This equals:\[x^{2k+2} + x^{2k+1}y + \, \cdots \, + xy^{2k+1} + y^{2k+2}\]

Step 4: Use Inductive Hypothesis

According to the inductive hypothesis for \( n = k \), we have:\[x^{2k} + x^{2k-1}y + \, \cdots \, + xy^{2k-1} + y^{2k} = \frac{x^{2k+1} - y^{2k+1}}{x-y}\]Adding the next terms on both sides, we get:\[x^{2k+2} + x^{2k+1}y + \, \cdots \, + xy^{2k+1} + y^{2k+2} = x^{2k} + x^{2k-1}y + \, \cdots \, + xy^{2k-1} + y^{2k} + x^{2k+1} + y^{2k+1}\]

Step 5: Simplify Using Algebra

We know that:\[(x^{2k+1} - y^{2k+1}) + (x^{2k+1} + y^{2k+1}) = x^{2k+2} - y^{2k+2}\]Thus, by the inductive hypothesis:\[\frac{x^{2k+3} - y^{2k+3}}{x-y}\] This matches the form of the original equation for \( n = k+1 \).

Step 6: Conclusion of Induction

Since both the base case and the inductive step are verified, by the principle of mathematical induction, the given statement is true for all positive integers \( n \).

Key Concepts

Base Case

When starting a proof by mathematical induction, it is crucial to establish the base case. This is the foundation of the entire proof process. The base case involves checking the statement for the smallest possible value of the variable, usually the integer 1. In our exercise, we substitute 1 for \( n \), giving us the equation:\[x^2 + xy + y^2 = \frac{x^3 - y^3}{x - y}\]By performing algebraic manipulations, we expand the right-hand side:\[\frac{(x-y)(x^2+xy+y^2)}{x-y} = x^2 + xy + y^2\]You'll notice that both sides of the equation turn out to be the same. This means that the base case holds true, confirming the expression is correct for \( n=1 \). Validating the base case is like planting the first seed in a garden – it's the initial step that allows further growth (or, in this context, proof).

Inductive Step

After establishing the base case, the next step is to perform the inductive step. This is similar to linking a chain, where each link must securely connect to the next.

• Assume the statement is true for a particular case \( n=k \).
• The assumption is: \( x^{2k} + x^{2k-1}y + \cdots + xy^{2k-1} + y^{2k} = \frac{x^{2k+1} - y^{2k+1}}{x-y} \).

This assumption is known as the inductive hypothesis. With it, we aim to prove that the statement holds for \( n = k+1 \). This means showing:\[x^{2(k+1)} + x^{2(k+1)-1}y + \cdots + xy^{2(k+1)-1} + y^{2(k+1)}\]To successfully connect what we know for \( n=k \) to what needs to be shown for \( n=k+1 \), we slightly modify the hypothesis by adding appropriate terms and using algebraic expansion.

Algebra

Algebra is the tool that we repeatedly use in proving statements by induction, especially during the inductive step. When working through the details, such as simplifying terms, factorizing expressions, and rearranging equations, algebra becomes invaluable.In the present problem, after the inductive hypothesis is stated, algebra comes into play to show equivalence when \( n=k+1 \). Adding and simplifying terms involves algebraic manipulation like this:\[(x^{2k+1} - y^{2k+1}) + (x^{2k+1} + y^{2k+1}) = x^{2k+2} - y^{2k+2}\]This algebraic step simplifies the process of transiting from the hypothesis to the requirement for \( n=k+1 \). Finally, the expression ends up exactly matching the form given in the original statement for \( n=k+1 \). This confirms our induction is successful, proving the statement true for all positive integers \( n \). Thus, algebra is the engine driving the logic behind induction forward.