Problem 20 - Calculus

Question

Verify that \(1^{3}+2^{3}+\cdots+n^{3}\) is \(f_{n}=\frac{1}{4} n^{2}(n+1)^{2}\) by checking \(f_{1}\) and \(f_{x}-f_{n-1}\). The text has a proof without words.

Step-by-Step Solution

Answer

The formula is verified for all \( n \).

Step 1: Verify the Base Case

First, check if the given formula holds for when \( n = 1 \). According to the problem, \( 1^3 = f_1 \). Substitute \( n = 1 \) in the formula: \[f_{1} = \frac{1}{4} \cdot 1^2 \cdot (1+1)^2 = \frac{1}{4} \cdot 1 \cdot 4 = 1\]Thus, the formula holds for the base case, \( n = 1 \).

Step 2: Express the Difference

Next, compute \( f_n - f_{n-1} \) to show the step from \( n-1 \) to \( n \). Substitute the given formula for both: \[f_n = \frac{1}{4}n^2(n+1)^2\]\[f_{n-1} = \frac{1}{4}(n-1)^2(n)^2\]

Step 3: Expand the Formulas

Expand both expressions obtained for \( f_n \) and \( f_{n-1} \). For \( f_n \):\[f_n = \frac{1}{4} n^2 (n^2 + 2n + 1)\]For \( f_{n-1} \): \[f_{n-1} = \frac{1}{4} (n^2 - 2n + 1)n^2 = \frac{1}{4} (n^4 - 2n^3 + n^2)\]

Step 4: Compute the Differences

Now, compute \( f_n - f_{n-1} \): \[f_n - f_{n-1} = \frac{1}{4} n^2 (n^2 + 2n + 1) - \frac{1}{4} (n^4 - 2n^3 + n^2)\]Simplify:\[= \frac{1}{4} \left(n^4 + 2n^3 + n^2 - n^4 + 2n^3 - n^2\right)\]\[= \frac{1}{4} \cdot 4n^3 = n^3\]It simplifies to \( n^3 \), thus verifying that the difference is indeed the cube of \( n \).

Step 5: Conclusion

Since both the base case and the inductive step have been verified, the given formula \( f_n = \frac{1}{4} n^{2}(n+1)^{2} \) correctly calculates the sum of cubes up to \( n \).

Key Concepts

Sum of Cubes

The sum of cubes refers to the expression when we add up the cubes of the first few natural numbers. Mathematically, it is represented as:

• \( 1^3, 2^3, 3^3, \ldots, n^3 \)

If you compute this series, you will notice a fascinating pattern. The sum of these cubes for any number \( n \) can be neatly calculated using a formula, which is:

• \( f_n = \frac{1}{4}n^2(n+1)^2 \)

What makes this formula even more amazing is its simplicity and the elegance with which it encapsulates the growth of cubic numbers. By using this formula, instead of adding each cube individually, we can derive the sum directly. This is not only efficient, but it also highlights a beautiful symmetry in numbers.

Inductive Step

The inductive step is a crucial part of mathematical induction. After establishing the base case, it demonstrates that if a formula holds for some integer \( n \), it also holds for \( n+1 \).In our exercise, we need to show that the formula:
\( f_{n} = \frac{1}{4}n^2(n+1)^2 \)
leads to a difference of \( n^3 \) when we compute
<\( f_{n} - f_{n-1} \). This difference confirms that each cube is accounted for in the total sum.By expanding the expression for both \( f_{n} \) and \( f_{n-1} \) and calculating their difference, we derive:

• \( f_n - f_{n-1} = n^3 \)

This establishes that every step in moving from \( n \) to \( n + 1 \) accommodates the next cube in the sequence. Thus, the formula holds throughout the process.

Base Case

The base case is the starting block of a proof by induction. It involves confirming that a given statement is true for the initial step, usually where \( n = 1 \). This provides the foundation of your inductive proof.In our sum of cubes problem, we begin by checking if the formula holds when \( n = 1 \). Calculating the cube of 1 gives:

• \( 1^3 = 1 \)

We then substitute \( n = 1 \) into the formula to find \( f_1 \):

• \( f_1 = \frac{1}{4} imes 1^2 imes (1+1)^2 = 1 \)

The results align perfectly, ensuring that the formula is accurate for the very first cube. Thus, the base case holds true, paving the way for the rest of the induction to proceed.

Proof by Induction

Proof by induction is a powerful method to establish that a given statement is true for all natural numbers. It follows two main steps: the base case and the inductive step.First, you validate the base case, demonstrating the problem's truth at a starting point, such as when \( n=1 \).

• In our example, we showed that \( 1^3 = 1 \) matched with the formula \( f_1 \).

Then, the inductive step assumes that the proposition is true for an arbitrary number \( n \), and under this assumption, shows that it must also be true for \( n+1 \).

• We computed \( f_n - f_{n-1} = n^3 \), confirming the formula extends to \( n+1 \).

These steps collaboratively reinforce the proof's accuracy. Base and inductive steps logically linked ensure that any conclusion drawn holds for any value of \( n \), allowing us to claim that the formula fits universally. This structure makes induction a systematic approach to tackle infinite cases.