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🁢 Mathematical Induction Lab

Why knocking over just one domino makes
them all fall in sequence

To prove that "1+2+⋯+n = n(n+1)/2" holds for every natural number n, you cannot check them one by one because there are infinitely many natural numbers. Instead, it is enough to show just two things — ① the first domino (n=1) falls, and ② whenever one domino falls, the very next domino must also fall. Once these two facts are established, all the dominoes will fall even though the chain continues forever.

k to check in the inductive step3
① Base case — Does the first domino fall?
When n = 1
Left side: 1
Right side: 1(1+1)/2 = 1
→ It holds! ✅
② Inductive step — Assume it holds when n = k
③ Then, does it also hold when n = k+1?

Why are these two steps enough? In ①, we confirmed that the domino for n=1 (the first domino) falls. In ②→③, we confirmed that "if the kth domino falls, the (k+1)th domino must also fall."

Then, because domino 1 falls, domino 2 falls; because domino 2 falls, domino 3 falls; and so on… Since this chain continues without breaking, we can conclude that the statement holds for every natural number n. Move the k slider and check that the same pattern repeats no matter which point in the chain you examine.

🁢 Mathematical Induction Quiz

Question 1/3 · Correct 0