proof by mathmatical induction that the sum of the first n natural numbers is equal n(n+1)/2

It's true for n = 1. Assume that it is true for some n. Then the sum of the first n+1 natural integers can be obtained by dding the last number n+1 to n(n+1)/2. So, the assumption that it is true for n leads to the conclusion that for n+1 the sum must be:

n+1 + n(n+1)/2.

If the formula is correct for n+1 also, then this must be the same as:

(n+1)(n+2)/2.

Expanding out the last bracket gives:

(n+1)(n+2)/2 = (n+1)*n/2 + (n+1)*2/2 =

n+1 + n(n+1)/2.

Thank you Count Iblis

To prove that the sum of the first n natural numbers is equal to n(n+1)/2 using mathematical induction, follow these steps:

Step 1: Base case
The base case is when n = 1. Show that the formula holds true when n = 1:
1 = 1(1+1)/2
1 = 2/2
1 = 1
Since the base case is true, let's move on to the next step.

Step 2: Assume the formula is true for n = k
Assume that the formula holds true when n = k. This is called the induction hypothesis. So, we assume that the sum of the first k natural numbers is given by k(k+1)/2.

Step 3: Show that the formula holds true when n = k+1
We need to prove that the formula also holds true for n = k+1. This means proving that the sum of the first k+1 natural numbers is given by (k+1)((k+1)+1)/2.

For n = k+1, the sum of the first k+1 natural numbers is:
1 + 2 + 3 + ... + k + (k+1) = (k(k+1)/2) + (k+1) [using the induction hypothesis]

To simplify further, we can multiply (k+1) to each term in the parentheses:
= (k(k+1) + 2(k+1))/2
= (k^2 + k + 2k + 2)/2
= (k^2 + 3k + 2)/2
= (k+2)(k+1)/2

Since (k+2)(k+1)/2 is equal to (k+1)((k+1)+1)/2, we have shown that the formula is also true for n = k+1.

Step 4: Conclude by mathematical induction
By completing both steps 1-3, we have proven that if the formula is true for any value of n (in our base case n=1), and if the formula is true for n = k, then it must also hold true for n = k+1. Hence, by mathematical induction, the formula is true for all natural numbers n.

Therefore, the sum of the first n natural numbers is equal to n(n+1)/2.