Problem 9 - Algebra and Trigonometry

Question

Use mathematical induction to prove that the formula is true for all natural numbers n. $$2^{3}+4^{3}+6^{3}+\cdots+(2 n)^{3}=2 n^{2}(n+1)^{2}$$

Step-by-Step Solution

Answer

The formula is true for all natural numbers by induction.

Step 1: Base Case

First, we check if the formula holds for the smallest natural number, typically \( n = 1 \). Substituting \( n = 1 \) into the formula's left side, we get \[(2 \times 1)^3 = 2^3 = 8.\]For the right side of the formula, substituting \( n = 1 \), we have \[2 \times 1^2 \times (1+1)^2 = 2 \times 1 \times 4 = 8.\]Both sides are equal, so the base case holds true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some arbitrary positive integer \( k \). Therefore, assume that \[2^3 + 4^3 + 6^3 + \cdots + (2k)^3 = 2k^2(k+1)^2.\] This assumption will be used to prove that the statement holds for \( k+1 \).

Step 3: Inductive Step

We need to prove that if the statement holds for \( n = k \), then it also holds for \( n = k+1 \). Consider the left-hand side of the equation at \( n = k+1 \), which is:\[2^3 + 4^3 + 6^3 + \cdots + (2k)^3 + (2(k+1))^3.\]Using the inductive hypothesis, this becomes:\[2k^2(k+1)^2 + (2(k+1))^3.\]Calculate \((2(k+1))^3:\)\[(2(k+1))^3 = 8(k+1)^3.\]

Step 4: Simplify and Prove for n = k+1

Substitute the expression for \((2(k+1))^3\) back into the equation:\[2k^2(k+1)^2 + 8(k+1)^3.\]Factor \((k+1)^2\) out of both terms:\[(k+1)^2 (2k^2 + 8(k+1)) = (k+1)^2 (2k^2 + 8k + 8).\]Simplify the expression inside the brackets:\[= (k+1)^2 (2k^2 + 8k + 8) = (k+1)^2 (2(k+2)^2).\]This simplifies further to:\[2(k+1)^2(k+2)^2,\]which matches the right-hand side for \( n = k+1 \). This completes the induction step.

Key Concepts

Natural Numbers

Natural numbers are the set of positive integers starting from 1 that we commonly use in mathematics for counting and ordering. They are represented as \( \{1, 2, 3, 4, \ldots\} \).
These numbers do not include zero, fractions, or negative numbers. In the context of mathematical induction, natural numbers form the foundation upon which we demonstrate a statement or formula applies to all possible cases.
When using induction, we always start with natural numbers because they provide a sequential path to assume and prove our statement.

Base Case

The base case is the initial step in mathematical induction, where we verify that the statement holds true for the smallest natural number. This smallest number is typically \( n = 1 \) unless stated otherwise.

• By checking the base case, we establish a starting point for our induction.
• If the statement is true for this foundational case, it sets the precedent for successive steps.

In our exercise, we verified the formula for \( n = 1 \), confirming that both sides equaled 8.
This successful verification of the base case forms the foundation upon which we build the inductive hypothesis and subsequent steps.

Inductive Hypothesis

The inductive hypothesis is a critical assumption made in the middle of a mathematical induction proof. Here, we assume that the statement is true for an arbitrary natural number \( k \).
This assumption allows us to test and demonstrate that if the statement holds for \( n = k \), it must also hold for the next number, \( n = k+1 \).

• The inductive hypothesis acts as a bridge, connecting the base case and the inductive step.
• By logically assuming this step, we can perform the necessary algebraic manipulations confidently.

In our exercise, the assumption was made that the formula holds for \( n = k \), forming the groundwork needed to prove for \( n = k+1 \).

Inductive Step

The inductive step is where the magic of mathematical induction happens. After establishing the base case and making an inductive hypothesis, the inductive step involves proving that if the statement is true for \( n = k \), then it is true for \( n = k+1 \) as well.

• This step involves algebraic manipulation and logical deduction, using the hypothesis as a base.
• In the provided exercise, this required calculating and simplifying expressions involving \( (k+1) \).
• We decomposed the equation \( 2^3 + 4^3 + \cdots + (2(k+1))^3 \) using our assumption, and ensured the statement holds for the next natural number.

Successfully demonstrating the inductive step confirms the validity of the formula for all natural numbers following the base case. It closes the loop, reinforcing our confidence in the proof.