Let's work through the problem step-by-step.
1. Base Case: Assume the statement is true for some [tex]\( n = k \)[/tex].
Let [tex]\( d(k) \)[/tex] represent the number of dots when [tex]\( n = k \)[/tex].
2. Inductive Step: We need to prove that if the statement holds for [tex]\( n = k \)[/tex], then it must also hold for [tex]\( n = k+1 \)[/tex].
According to our assumption, [tex]\( d(k) \)[/tex] is the total number of dots when [tex]\( n = k \)[/tex].
3. Incrementing [tex]\( n \)[/tex]: Suppose we increase [tex]\( n \)[/tex] by 1, that is, we examine the case [tex]\( n = k+1 \)[/tex].
4. Expression for [tex]\( n = k+1 \)[/tex]: When [tex]\( n \)[/tex] becomes [tex]\( k + 1 \)[/tex], it means we are adding an additional level with [tex]\( k + 1 \)[/tex] dots to our total dot count.
5. Inductive Hypothesis: We assume
[tex]\[
d(k+1) = d(k) + (k + 1)
\][/tex]
Here, [tex]\( d(k) \)[/tex] is the number of dots if [tex]\( n = k \)[/tex] and adding [tex]\( k + 1 \)[/tex] dots for the next level.
6. Combining Information:
Let’s denote the number of dots at [tex]\( n = k+1 \)[/tex] by [tex]\( d(k+1) \)[/tex]. According to our induction step, we have:
[tex]\[
d(k+1) = d(k) + (k + 1)
\][/tex]
This shows that the number of dots for [tex]\( n = k+1 \)[/tex] is built on the number of dots for [tex]\( n = k \)[/tex] by adding [tex]\( k+1 \)[/tex] more dots.
7. Conclusion:
We’ve shown that if the statement holds for [tex]\( n = k \)[/tex], i.e., [tex]\( d(k) \)[/tex] dots, then it also holds for [tex]\( n = k+1 \)[/tex], i.e., [tex]\( d(k+1) = d(k) + (k + 1) \)[/tex].
By mathematical induction, since the statement is true for [tex]\( n=k \)[/tex] and leads directly to being true for [tex]\( n=k+1 \)[/tex], we conclude that the statement is true for all natural numbers [tex]\( n \)[/tex].
