Problem 11 - 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 proven true for all natural numbers \(n\) by mathematical induction.

Step 1: Understand the Problem

We need to prove that the sum of cubes of even numbers from 2 to \((2n)^3\) is equal to \(2n^2(n+1)^2\) for all natural numbers \(n\). This is a problem that can be solved using mathematical induction.

Step 2: Base Case

First, we test the formula for the smallest natural number, \(n = 1\):\[2^3 = 8\]The right-hand side is:\[2 \times 1^2 \times (1+1)^2 = 2 \times 1 \times 4 = 8\]Since both sides are equal, the base case holds.

Step 3: Inductive Hypothesis

Assume that the formula holds for some arbitrary natural number \(k\), i.e.,\[2^3 + 4^3 + 6^3 + \.\.\.+ (2k)^3 = 2k^2(k+1)^2\]This is our inductive hypothesis.

Step 4: Inductive Step

We need to prove that the formula holds for \(n = k + 1\). According to our hypothesis:\[2^3 + 4^3 + 6^3 + \.\.\. + (2k)^3 + (2(k+1))^3 = 2(k+1)^2(k+2)^2\]Express the left-hand side using the hypothesis:\[LHS = 2k^2(k+1)^2 + (2(k+1))^3\]Substitute and simplify:\[= 2k^2(k+1)^2 + 8(k+1)^3\]Factor out \((k+1)^2\):\[= (k+1)^2[2k^2 + 8(k+1)]\]Multiply terms inside the bracket:\[= (k+1)^2[2k^2 + 8k + 8]\]\[= (k+1)^2[2(k+2)^2]\]This simplifies to the RHS:\[2(k+1)^2(k+2)^2\]Thus, the formula holds for \(k+1\).

Step 5: Conclusion

Since the base case holds and the inductive step proves that if the formula holds for \(k\), then it also holds for \(k+1\), by induction, the formula is true for all natural numbers \(n\).

Key Concepts

Sum of Cubes

The sum of cubes refers to the result obtained when you cube a sequence of numbers and add those cubes together. In the context of the problem we have, specifically the sum of cubes of even numbers, we are cubing and then adding even natural numbers, such as 2, 4, 6, and so on, up to a certain number. To provide an example, consider the sequence 2, 4, and 6. The sum of cubes for these numbers would be:

• \(2^3 = 8\)
• \(4^3 = 64\)
• \(6^3 = 216\)

Adding these values gives \(8 + 64 + 216 = 288\). In mathematical induction, our goal is to not only verify this sum for a small number of even numbers but for all even numbers up to \(2n\), and verify that it equals \(2n^2(n+1)^2\). This formula provides a quick way to evaluate the sum without calculating each cube individually.

Even Numbers

Even numbers are integers that can be exactly divided by 2, meaning there is no remainder. The sequence of even numbers starts from 2, 4, 6, and continues indefinitely. In this exercise, we focus on cubing and summing even numbers up to a specific number \(2n\). Even numbers are crucial here because they form the basis of our sequence. We cube each even number and then sum these cubes to form one side of the equation in our inductive proof. Understanding that the series only includes even numbers is key to setting up the inductive proof correctly, as each step and formula revolves around this sequence.

Natural Numbers

Natural numbers are a subset of numbers starting from 1 and increasing by 1 indefinitely. These are sometimes referred to as counting numbers and do not include negative numbers, fractions, or decimals.In the context of our task, natural numbers \(n\) represent the number of terms—or pairs of terms, since we are dealing with evens—in our sequence, from \(2\) to \(2n\). For example, if \(n = 3\), the natural numbers we consider in terms of even numbers would be \(2, 4,\) and \(6\).The problem statement utilizes natural numbers to outline the limits up to which the even numbers are cubed and summed, ensuring we can generalize the solution across an infinite sequence when demonstrating it with mathematical induction.

Inductive Proof

An inductive proof is a mathematical method used to prove a statement is true for all natural numbers. It generally consists of two main steps: the base case and the inductive step.

• Base Case: Verified for the smallest value of the natural number, often \(n=1\). In our problem, the base case shows that both sides of the equation are equal when \(n=1\).
• Inductive Step: We assume the statement holds for an arbitrary natural number \(k\) and then prove it also holds for \(k+1\). This involves expressing both sides of the equation for \(n=k\), modifying the left by including an additional term \((2(k+1))^3\), and showing it equates to the formula devised for \(n=k+1\).

By completing both, one establishes the statement's truth for all natural numbers, which closes the proof under the principle of mathematical induction.