Proof By Induction: Sum Of Arithmetic Series

Let's dive into a fascinating mathematical concept: proof by induction. Today, we're going to use it to demonstrate a specific formula for the sum of an arithmetic series. Specifically, we aim to prove that if the statement $S_k: 19+38+57+\ldots+19 k=\frac{19 k(k+1)}{2}$ is true for some natural number $k$, then the statement $S_{k+1}: 19+38+57+\ldots+19 k+19(k+1)=\frac{19(k+1)(k+2)}{2}$ is also true. We will also rewrite the left side of $S_{k+1}$ to express it as $19+38+57+\ldots+19 k+19(k+1)=\square+19(k+1)$, identifying what should go in the blank.

Understanding Mathematical Induction

Mathematical induction is a powerful technique used to prove statements that hold for all natural numbers. It's like setting up a line of dominoes; if you can knock down the first domino, and if each domino knocks down the next, then you can be sure all the dominoes will fall. There are typically two main steps:

1. Base Case: Show that the statement is true for the smallest natural number (usually $k = 1$).
2. Inductive Step: Assume the statement is true for some arbitrary natural number $k$ (the inductive hypothesis), and then prove that it must also be true for $k + 1$.

If we can successfully complete these two steps, the principle of mathematical induction tells us that the statement is true for all natural numbers.

Defining $S_k$ and $S_{k+1}$

We are given the statement $S_k: 19+38+57+\ldots+19 k=\frac{19 k(k+1)}{2}$. This statement proposes a formula for the sum of the first $k$ terms of an arithmetic series where each term is a multiple of 19. The series starts with 19, then 38, then 57, and so on, up to $19k$.

Our goal is to prove $S_{k+1}: 19+38+57+\ldots+19 k+19(k+1)=\frac{19(k+1)(k+2)}{2}$. This statement extends the sum to the first $k+1$ terms, including the term $19(k+1)$. We want to show that if $S_k$ is true, then $S_{k+1}$ must also be true.

Rewriting the Left Side of $S_{k+1}$

Before we dive into the full induction proof, let's focus on rewriting the left side of the $S_{k+1}$ equation. This will make the inductive step clearer. The left side is:

$19+38+57+\ldots+19 k+19(k+1)$

We want to express this in the form $\square + 19(k+1)$. Notice that the terms $19+38+57+\ldots+19k$ are exactly the terms that $S_k$ describes. Therefore, we can directly replace these terms with the right side of the $S_k$ equation, which is $\frac{19 k(k+1)}{2}$.

So, we have:

$19+38+57+\ldots+19 k+19(k+1) = \frac{19 k(k+1)}{2} + 19(k+1)$

This means that the blank $\square$ should be filled with $\frac{19 k(k+1)}{2}$. This rewriting step is crucial because it connects $S_{k+1}$ back to $S_k$, which is the foundation of the inductive step.

Proof of $S_{k+1}$ by Induction

Now, let's proceed with the complete proof by induction.

1. Base Case

We need to show that $S_k$ is true for the smallest natural number, which is $k=1$. So we want to check if $S_1$ is true.

$S_1: 19(1) = \frac{19(1)(1+1)}{2}$ $19 = \frac{19(2)}{2}$ $19 = 19$

The base case holds true.

2. Inductive Step

Assume $S_k$ is true for some arbitrary natural number $k$. This is our inductive hypothesis:

$S_k: 19+38+57+\ldots+19 k=\frac{19 k(k+1)}{2}$

We need to prove that $S_{k+1}$ is also true, which is:

$S_{k+1}: 19+38+57+\ldots+19 k+19(k+1)=\frac{19(k+1)(k+2)}{2}$

Starting with the left side of $S_{k+1}$, we have:

$19+38+57+\ldots+19 k+19(k+1)$

Using our rewriting from earlier, we substitute the sum of the first $k$ terms with $\frac{19 k(k+1)}{2}$ based on our inductive hypothesis:

$\frac{19 k(k+1)}{2} + 19(k+1)$

Now, we want to manipulate this expression to match the right side of $S_{k+1}$, which is $\frac{19(k+1)(k+2)}{2}$. Let's factor out the common term $19(k+1)$:

$19(k+1) \left( \frac{k}{2} + 1 \right)$

To combine the terms inside the parenthesis, we need a common denominator:

$19(k+1) \left( \frac{k}{2} + \frac{2}{2} \right)$

$19(k+1) \left( \frac{k+2}{2} \right)$

Now, we can rewrite this as:

$\frac{19(k+1)(k+2)}{2}$

This is exactly the right side of $S_{k+1}$. Therefore, we have shown that if $S_k$ is true, then $S_{k+1}$ is also true.

Conclusion

By successfully demonstrating both the base case and the inductive step, we have proven by mathematical induction that the formula $S_k: 19+38+57+\ldots+19 k=\frac{19 k(k+1)}{2}$ holds for all natural numbers $k$. We also showed that the left side of $S_{k+1}$ can be rewritten as $\frac{19 k(k+1)}{2} + 19(k+1)$. This showcases the power and elegance of mathematical induction in proving statements about sequences and series.

For further reading on mathematical induction, check out this resource on Wikipedia.