Let's prove that the given statement is true for all natural numbers [tex]\( n \)[/tex] using the method of mathematical induction.
Statement: For all natural numbers [tex]\( n \)[/tex], [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex], where [tex]\( d(n) \)[/tex] is the total number of dots.
Step 1: Base Case
First, we need to verify the statement for the smallest natural number, which is [tex]\( n = 1 \)[/tex].
When [tex]\( n = 1 \)[/tex]:
[tex]\[ d(1) = \frac{1(1+1)}{2} = \frac{2}{2} = 1 \][/tex]
Since the total number of dots when [tex]\( n = 1 \)[/tex] is indeed 1, the base case holds.
Step 2: Induction Hypothesis
Assume that the statement is true for some arbitrary natural number [tex]\( k \)[/tex]. That is, assume:
[tex]\[ d(k) = \frac{k(k+1)}{2} \][/tex]
Step 3: Induction Step
We need to prove that the statement is true for [tex]\( n = k + 1 \)[/tex]. That is, we need to show:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
According to the given hint, the total number of dots at [tex]\( n = k+1 \)[/tex] can be expressed in terms of [tex]\( d(k) \)[/tex] as follows:
[tex]\[ d(k+1) = d(k) + (k+1) \][/tex]
Using our induction hypothesis [tex]\( d(k) = \frac{k(k+1)}{2} \)[/tex], we can substitute this into the equation:
[tex]\[ d(k+1) = \frac{k(k+1)}{2} + (k+1) \][/tex]
To simplify this, first factor out [tex]\( (k+1) \)[/tex] from the equation:
[tex]\[ d(k+1) = (k+1) \left( \frac{k}{2} + 1 \right) \][/tex]
Combine the fractions inside the parentheses:
[tex]\[ d(k+1) = (k+1) \left( \frac{k + 2}{2} \right) \][/tex]
Now, multiply [tex]\( (k+1) \)[/tex] by [tex]\( \frac{k+2}{2} \)[/tex]:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
Therefore:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
We have shown that if the statement is true for [tex]\( n = k \)[/tex], then it must also be true for [tex]\( n = k+1 \)[/tex].
Conclusion
By the principle of mathematical induction, since the base case holds and the induction step has been proven, the statement is true for all natural numbers [tex]\( n \)[/tex]:
[tex]\[ d(n) = \frac{n(n+1)}{2} \][/tex]
Thus, the original statement has been proven for all natural numbers.
