Problem 43 - Discrete Mathematics with Applications

Question

Prove the binomial theorem, using mathematical induction.

Step-by-Step Solution

Answer

We proved the binomial theorem using mathematical induction in 2 steps: 1. Base Case (n = 0): We showed that \((a+b)^0=\sum_{k=0}^0 {0 \choose k} a^{0-k}b^k\) by proving both sides equal 1. 2. Inductive Step (n = p and n = p + 1): We assumed the theorem holds for n = p and then showed it also holds for n = p + 1 by combining the two summations and applying the identity \({n \choose k} + {n \choose k-1} = {n+1 \choose k}\), thus proving the binomial theorem true for all natural numbers n.

Step 1: Prove Base Case (n = 0)

In this step, we want to show that the binomial theorem holds for n=0. According to the theorem, \[(a+b)^0=\sum_{k=0}^0 {0 \choose k} a^{0-k}b^k\] Now, we know that any non-zero number to the power of 0 is equal to 1, so \((a+b)^0 = 1\). As for the right-hand side of the equation, there's only one term in the summation since k=0 is the only value in the range, so we have: \[\sum_{k=0}^0 {0 \choose k} a^{0-k}b^k = {0 \choose 0}a^0b^0\] By definition, \({n \choose n} = {0 \choose 0} = 1\). Also, \(a^0 = b^0 = 1\). Therefore, \[{0 \choose 0}a^0b^0 = 1 * 1 * 1 = 1\] Since both the left-hand side and the right-hand side of the equation equal 1, we have proved the base case: \[(a+b)^0=\sum_{k=0}^0 {0 \choose k} a^{0-k}b^k\]

Step 2: Inductive Step (n = p and n = p + 1)

In this step, we want to assume the binomial theorem is true for n=p and show that it also holds for n=p+1. 1. Inductive Hypothesis: Assume the theorem is true for n=p, i.e., \[(a+b)^p = \sum_{k=0}^p {p \choose k} a^{p-k} b^k\] 2. Now, let's show that it also holds for n=p+1, \[(a+b)^{p+1} = \sum_{k=0}^{p+1} {p+1 \choose k} a^{p+1-k} b^k\] To do this, we'll start by expanding \((a+b)^{p+1}\). \[(a+b)^{p+1} = (a+b)(a+b)^p\] Now, using the inductive hypothesis, we can substitute the expression on the right-hand side of the equation: \[(a+b)^{p+1} = (a+b)\left(\sum_{k=0}^p {p \choose k} a^{p-k} b^k\right)\] Next, we distribute the (a+b) term: \[(a+b)^{p+1} = \sum_{k=0}^p {p \choose k} a^{p+1-k} b^k + \sum_{k=0}^p {p \choose k} a^{p-k} b^{k+1}\] Now, we rewrite the second summation with the index shifted by 1 (replace k with k-1): \[(a+b)^{p+1} = \sum_{k=0}^p {p \choose k} a^{p+1-k} b^k + \sum_{k=1}^{p+1} {p \choose k-1} a^{p+1-k} b^{k}\] We can combine the two summations into one: \[(a+b)^{p+1} = \sum_{k=0}^{p+1} \left({p \choose k} a^{p+1-k} b^k + {p \choose k-1} a^{p+1-k} b^{k}\right)\] Now, we'll use the identity for the binomial coefficients: \[{n \choose k} + {n \choose k-1} = {n+1 \choose k}\] Applying this identity, we get: \[(a+b)^{p+1} = \sum_{k=0}^{p+1} {p+1 \choose k} a^{p+1-k} b^k\] This is what we wanted to prove; the binomial theorem holds for n=p+1. Since we have proved both the base case and the inductive step, we can conclude that the binomial theorem is true for all natural numbers n, using mathematical induction.

Key Concepts

Mathematical Induction

Mathematical induction is like a domino effect for mathematics. It's a powerful technique used to prove that a statement is true for all natural numbers. To use it, you'll typically start with a base case and an inductive step. These two parts work together to prove that if a statement is true for one case, it must also be true for the next.

In simpler terms, mathematical induction logically connects each case to the next one, ensuring the truth of the statement continuously down the line. This method is particularly useful in proving statements that involve large or infinite sets of numbers, like the binomial theorem.

Base Case

The base case is the starting point in a proof using mathematical induction. You can think of it as the block laying the foundation for a series of falling dominoes. It's the simplest version of the statement, usually when the variable is zero or one.

In the binomial theorem's proof, the base case is when \(n = 0\). Here, we show that both sides of the equation transform into the value "1", confirming the theorem holds true. By establishing that the base case is accurate, we create a firm base to build upon with the inductive step.

Inductive Step

After proving the base case, the next part is the inductive step. This step involves two vital pieces:

• Assume the statement is true for some arbitrary case, let's call it \(n = p\) (this part is known as the inductive hypothesis).
• Show that if the statement is true for \(n = p\), it must also be true for \(n = p+1\).

In the example of the binomial theorem, the inductive step verifies that if the equation holds for \(n = p\), it must also hold for \(n = p+1\). Using logical steps and the identity of binomial coefficients, it seamlessly transitions from one case to the next, cementing the theorem across all natural numbers.

Binomial Coefficient

A binomial coefficient is a special kind of number that's crucial in the binomial theorem. It appears in expressions like \({n \choose k}\) and is used to represent the number of ways to choose \(k\) items from \(n\) items without regard to the order.

Mathematically, a binomial coefficient is calculated using the formula: \[{n \choose k} = \frac{n!}{k!(n-k)!}\]where \(!\) denotes factorial, which is the product of all positive integers up to that number.
Binomial coefficients are significant because they help break down the expression \((a+b)^n\) in the binomial theorem by identifying how many times each term should appear. They provide structure and clarity in expanding such expressions, essential to the theorem's proof and applications.