To prove that the sum of the first n natural numbers is equal to n times (n plus one) divided by two for all natural numbers, we use the method of mathematical induction.
For n = 1:
When n = 1, the left-hand side is 1
The right-hand side is 1 × (1 + 1) ÷ 2 = 1.
Since both sides are equal, the formula holds true for n=1.

Assume true for n = k:
Assume that for some natural number k, the formula holds.
That is,
1 + 2 + 3 + … + k = k × (k + 1) ÷ 2.
This is the induction hypothesis.

To prove for n = k + 1:
Now, consider the sum up to k+1:
1 + 2 + 3 + … + k + (k + 1)
By the induction hypothesis, the sum up to k is k × (k + 1) ÷ 2.
Adding k+1 gives:
[k × (k + 1) ÷ 2] + (k + 1)
Taking k+1 common:
= (k + 1) × [k ÷ 2 + 1]
= (k + 1) × [(k + 2) ÷ 2]
= (k + 1) × (k + 2) ÷ 2

Which is the same formula with n = k + 1

Since the formula is true for n = 1, and assuming it holds for n = k implies it holds for n = k + 1, by the principle of mathematical induction, the formula is true for all natural numbers n ≥ 1.

Hence proved.