Assume the statement is true for [tex]n = k[/tex]. Prove that it must be true for [tex]n = k + 1[/tex], thereby proving it true for all natural numbers [tex]n[/tex].

Hint: Since the total number of dots increases by [tex]n[/tex] each time, prove that [tex]d(k) + (k + 1) = d(k + 1)[/tex].

Is it true because it increases by one dot each time?



Answer :

To prove that the statement is true for all natural numbers \( n \), we will use the principle of mathematical induction.

### Step 1: Base Case
First, we need to confirm the base case. Let's consider \( n = 1 \).

For \( n = 1 \), the number of dots, \( d(1) \), is 1.

So, \( d(1) = 1 \).

### Step 2: Inductive Hypothesis
Assume that the statement is true for \( n = k \). That means,
[tex]\[ d(k) = \frac{k \times (k + 1)}{2} \][/tex]

### Step 3: Inductive Step
We need to show that if the statement holds for \( n = k \), then it also holds for \( n = k + 1 \).

We know that by the assumption for \( n = k \):
[tex]\[ d(k) = \frac{k \times (k + 1)}{2} \][/tex]

We need to prove:
[tex]\[ d(k+1) = d(k) + (k + 1) \][/tex]

Let's compute \( d(k+1) \) using the inductive hypothesis:
[tex]\[ d(k+1) = d(k) + (k + 1) = \frac{k \times (k + 1)}{2} + (k + 1) \][/tex]

Combine the terms over a common denominator:
[tex]\[ d(k+1) = \frac{k \times (k + 1)}{2} + \frac{2 \times (k + 1)}{2} = \frac{k \times (k + 1) + 2 \times (k + 1)}{2} = \frac{(k + 2) \times (k + 1)}{2} = \frac{(k + 1) \times (k + 2)}{2} \][/tex]

Since we have shown:
[tex]\[ d(k+1) = \frac{(k+1) \times (k+2)}{2} \][/tex]

This matches the pattern for triangular numbers, confirming our inductive step.

### Conclusion
By proving the base case for \( n = 1 \) and using the inductive step to show that \( d(k) \) being true implies \( d(k+1) \) is also true, we have proven by induction that the statement is true for all natural numbers \( n \).

Hence, the formula holds as demonstrated by the inductive proof. For instance, if \( k = 5 \):
[tex]\[ d(5) = \frac{5 \times (5 + 1)}{2} = 15 \][/tex]
[tex]\[ d(6) = d(5) + 6 = 15 + 6 = 21 \][/tex]
Thus,
[tex]\[ d(5) = 15 \quad \text{and} \quad d(6) = 21 \][/tex]