Problem 67 - Essential Calculus

Question

Prove each formula by mathematical induction. (You may need to review the method of proof by induction from a precalculus text.) (a) \(\sum_{i=1}^{n} 2 i=n(n+1)\) (b) \(\sum_{i=1}^{n} i^{3}=\frac{n^{2}(n+1)^{2}}{4}\)

Step-by-Step Solution

Answer

For both formulas, the Mathematical Induction method has been used to prove that these are true for n=1, assumed to hold true for n=k, and subsequently proven to be correct for n=k+1. Therefore, they are proven to hold for all values of positive integers.

Step 1: Proving the base case for first formula

We start with n = 1 for the first formula. Substitute n with 1 in \(\sum_{i=1}^{n} 2 i=n(n+1)\), and the equation becomes \(2=1*(1+1)\). Both sides of the equation equal to 2, so the formula is valid for n=1.

Step 2: Proving the inductive step for first formula

Assume that the formula is true for n=k. So, \(\sum_{i=1}^{k} 2 i=k(k+1)\). Now, we must prove it holds for n=k+1. Adding 2(k+1) to both sides: \(\sum_{i=1}^{k+1} 2 i = 2(k+1) + k(k+1) = (k+1)(k+2)\). This shows that, if the formula is true for n=k, then it's also true for n=k+1.

Step 3: Proving the base case for second formula

Again, we start with n = 1, this time for the second formula. Substitute n with 1 in \(\sum_{i=1}^{n} i^{3} = \frac{n^{2}(n+1)^{2}}{4}\), and the equation becomes \(1^3 = \frac{1^2(1+1)^{2}}{4}\). Both sides of the equation equal to 1, so the formula is valid for n=1.

Step 4: Proving the inductive step for second formula

Assume that the formula is true for n=k: \(\sum_{i=1}^{k} i^{3} = \frac{k^{2}(k+1)^{2}}{4}\). Now, we must prove it holds for n=k+1. Adding \((k+1)^3\) to both sides: \(\sum_{i=1}^{k+1} i^{3} = (k+1)^{3} + \frac{k^{2}(k+1)^{2}}{4} = \frac{(k+1)^{2}(k+2)^{2}}{4}\). Hence, if the formula is true for n=k, then it's also true for n=k+1.

Key Concepts

Inductive Step

The inductive step is a crucial part of proving any statement by mathematical induction. It involves two main tasks: the assumption and the extension.

First, assume that the formula or statement is true for a specific case, usually represented as \(n = k\). This is called the "inductive hypothesis."

For our specific example involving the formula \(\sum_{i=1}^{n} 2i = n(n+1)\), the assumption would mean believing that this holds true for \(n = k\), so \(\sum_{i=1}^{k} 2i = k(k+1)\).

• Next, use this assumption to demonstrate the formula holds for \(n = k+1\).
• You essentially need to "add one more step" to your sum, which involves adding the next term in the sequence to both sides of your equation.
• In our case, that means proving that \(\sum_{i=1}^{k+1} 2i = (k+1)(k+2)\).

By successfully completing the inductive step, you hence show that if it holds for one number, it holds for the next, thereby supporting the case that it works for all natural numbers within the domain.

Base Case

The base case is like the foundation of a building in the method of mathematical induction. It's the checking point that ensures the formula or statement holds true initially.

This step involves substituting the smallest number in your domain, usually \(n = 1\), into the formula. For the formula \(\sum_{i=1}^{n} 2i = n(n+1)\), we substitute \(n = 1\) and check:

• The left side becomes \(2 \cdot 1 = 2\).
• The right side becomes \(1(1+1) = 2\).

Both sides equal 2, confirming the base case for \(n=1\). Successfully proving the base case means you've started your induction on solid ground. This essential step guarantees that your formula does not merely work by accident for larger numbers but is truly valid right from the start.

Summation Formulas

Summation formulas are mathematical expressions that provide a shorthand way to calculate the sum of a series of numbers. Rather than add each number individually, these formulas allow you to compute the total sum with a simple equation.

There are various summation formulas, each tailored to specific types of sequences:

• Arithmetic series: The sum of an arithmetic sequence with the formula \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\), which is particularly useful for adding the first \(n\) positive integers.
• Geometric series: Useful for sums of numbers in a geometric sequence, such as \(a + ar + ar^2 + ... + ar^n\).
• Special cubes: For sequences where each term is a cube, like our formula \(\sum_{i=1}^{n} i^{3}=\frac{n^{2}(n+1)^{2}}{4}\).

Summation formulas not only simplify calculations but are also powerful tools for proving complex results. When applying mathematical induction, these formulas allow proof by demonstrating how each term contributes systematically to the sequence's sum.