To prove by induction that [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex], we first need to confirm that the statement is true for [tex]\( n = 1 \)[/tex].
### Base Case:
Let's start with [tex]\( n = 1 \)[/tex]:
When [tex]\( n = 1 \)[/tex], there is 1 dot. According to the formula [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex]:
[tex]\[
d(1) = \frac{1(1+1)}{2} = \frac{1 \times 2}{2} = \frac{2}{2} = 1
\][/tex]
We observe that the number of dots when [tex]\( n = 1 \)[/tex] is indeed 1, which matches our derived value using the formula. Therefore, the statement holds true for [tex]\( n = 1 \)[/tex].
Hence, the statement [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex] is true for the base case [tex]\( n = 1 \)[/tex].
