Final answer:
The sum of the first *n* natural numbers, 1+2+⋯+*n*, is proven to be equal to *n(n+1)/2* for all *n* ∈ ℕ.
Step-by-step explanation:
To prove this formula, we use mathematical induction. The base case, *n=1*, is trivial as 1 = 1(1+1)/2. Now, assume the formula holds for some arbitrary *k* ∈ ℕ, i.e., 1+2+⋯+*k* = *k(k+1)/2*. We need to show that it also holds for *k+1*.
The induction step involves adding *k+1* to both sides of the assumed formula. Starting with 1+2+⋯+*k*, when we add *k+1*, we get *k(k+1)/2 + (k+1)*. Factoring out *k+1*, we obtain (*k+1*)(*k/2 + 1*). Simplifying further, this expression equals (*k+1*)(*k+2*/2), which completes the induction step. Therefore, the formula holds for *k+1*, assuming it holds for *k*.
In conclusion, by establishing the base case and demonstrating the induction step, we prove that 1+2+⋯+*n* is equal to *n(n+1)/2* for all *n* ∈ ℕ. This proof method ensures the validity of the formula for all natural numbers.