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]
### 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]
