Problem 11 - Precalculus: Mathematics for Calculus

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 \(n\).

Step 1: Base Case

First, we need to check if the formula holds true for the first natural number, which is \(n = 1\). Substituting \(n = 1\) into both sides of the equation:- Left side: The sum consists of only one term here, which is \((2 \cdot 1)^{3} = 2^{3} = 8\).- Right side: \(2 \cdot 1^{2} \cdot (1 + 1)^{2} = 2 \cdot 1 \cdot 4 = 8\).Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Assume that the formula holds for \(n = k\), where \(k\) is an arbitrary natural number. This means we assume \[2^{3}+4^{3}+6^{3}+ar{\cdots}+(2k)^{3}=2k^{2}(k+1)^{2}\]is true.

Step 3: Inductive Step

We need to show the statement holds for \(n = k + 1\). This means proving: \[2^{3}+4^{3}+6^{3}+ar{\cdots}+(2k)^{3} + (2(k+1))^{3} = 2(k+1)^{2}((k+1)+1)^{2}\]Substitute the inductive hypothesis into the left side:\[2k^{2}(k+1)^{2} + (2(k+1))^{3}\]This simplifies to: \[2k^{2}(k+1)^{2} + 8(k+1)^{3}\]

Step 4: Simplify the Expression

Next, we simplify the expression \[2k^{2}(k+1)^{2} + 8(k+1)^{3}\]:Factor out \((k+1)^{2}\):\[(k+1)^{2}(2k^{2} + 8(k+1)) = (k+1)^{2}(2k^{2} + 8k + 8)\]Simplify the term inside the parentheses:\[(k+1)^{2}(2k^{2} + 8k + 8) = (k+1)^{2}((k+1)^{2})\] So, it simplifies to:\[2(k+1)^{2}((k+1)+1)^{2}\]Both sides are equal, thus the formula holds for \(n = k + 1\).

Step 5: Conclusion

Since the base case holds and the truth of the statement for \(n = k\) implies its truth for \(n = k+1\), by the principle of mathematical induction, the formula \[2^{3}+4^{3}+6^{3}+ar{\cdots}+(2n)^{3} = 2n^{2}(n+1)^{2}\]is true for all natural numbers \(n\).

Key Concepts

Natural Numbers

Natural numbers are the set of positive integers starting from 1 and continuing upwards:

• 1, 2, 3, 4, ...

They do not include zero, fractions, or negative numbers. These numbers are fundamentally essential in counting and ordering. Understanding natural numbers is a key step when learning about topics like mathematical induction and arithmetic sequences, as they form the basis of many problems and solutions in mathematics. The exercise here requires proving a formula’s validity across all natural numbers using a specific methodology.

Inductive Hypothesis

The inductive hypothesis is a critical step in proving statements using mathematical induction. Here, it involves assuming that a statement holds true for some arbitrary natural number, say, "k". This assumption is not random but is a temporary step meant to bridge the base case with the more significant claim across all natural numbers. In the problem at hand, our inductive hypothesis is:

• Assume the formula \[2^3+4^3+6^3+\cdots+(2k)^3 = 2k^2(k+1)^2\] holds.

This hypothesis sets the stage for us to show that if this holds for one number, it leads to the possibility of it being true for the next number in sequence, ultimately proving validity for all natural numbers.

Base Case

The base case is the initial step in mathematical induction. It demonstrates that the statement or formula to be proven is true for the initial value in the sequence of natural numbers. Usually, it involves proving the formula for \(n = 1\). For this problem, when we replace \(n\) with 1, both sides of our formula yield 8, thereby confirming that:

• \[(2 \cdot 1)^3 = 8\]
• \[2 \cdot 1^2 \cdot (1+1)^2 = 8\]

Since both sides of the equation are equal, our base case is established. With this foundation, we can confidently move to the next steps required by mathematical induction.

Inductive Step

The inductive step in mathematical induction ensures that if a statement holds true for an arbitrary natural number \(k\), it must also be true for the next integer, \(k + 1\). This step transforms the hypothesis into a full-fledged proof structure. In handling our current exercise:

• We start by adding the next term, \((2(k+1))^3\), to both sides of our formula.
• We use the inductive hypothesis that supports our claim for \(n = k\).
• Finally, the expression is simplified to show that both sides of the equation for \(n = k+1\) are equal.

Thus, by proving that from \(n = k\) follows \(n = k + 1\), using the established base case, we confirm that our formula holds for all natural numbers. This conclusive step is what makes mathematical induction such a powerful tool in proving statements formed with natural numbers.