We are tasked with proving by induction that the following statement holds for all natural numbers $k \in \mathbb{N}$:

$\sum_{i=0}^{k} i = \frac{k(k+1)}{2}$

Step 1: Base Case

First, check if the formula holds for $k = 0$.

$\sum_{i=0}^{0} i = 0$ The right-hand side of the equation when $k = 0$ is:

$\frac{0(0+1)}{2} = \frac{0}{2} = 0$

Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Assume the formula holds for some arbitrary $k = n$, i.e.,

$\sum_{i=0}^{n} i = \frac{n(n+1)}{2}$

This is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove that if the statement holds for $k = n$, then it must also hold for $k = n + 1$. That is, we need to show:

$\sum_{i=0}^{n+1} i = \frac{(n+1)(n+2)}{2}$

Start with the left-hand side of the equation:

$\sum_{i=0}^{n+1} i = \sum_{i=0}^{n} i + (n+1)$

Using the inductive hypothesis, substitute $\sum_{i=0}^{n} i$ with $\frac{n(n+1)}{2}$:

$\sum_{i=0}^{n+1} i = \frac{n(n+1)}{2} + (n+1)$

Now, simplify the right-hand side:

$= \frac{n(n+1)}{2} + \frac{2(n+1)}{2}$ $= \frac{n(n+1) + 2(n+1)}{2}$ $= \frac{(n+1)(n+2)}{2}$

Thus, we have shown that:

$\sum_{i=0}^{n+1} i = \frac{(n+1)(n+2)}{2}$

Step 4: Conclusion

Since the formula holds for $k = 0$ (base case) and the inductive step has been proven, by mathematical induction, the formula is true for all $k \in \mathbb{N}$:

$\sum_{i=0}^{k} i = \frac{k(k+1)}{2}$

Would you like further details or explanations on this? Here are five related questions you might find interesting:

1. How does mathematical induction work in other types of summation formulas?
2. Can we use a similar approach to prove formulas involving higher powers of $i$ (e.g., $\sum_{i=0}^{k} i^2$)?
3. How can we derive this summation formula without induction (e.g., using a direct combinatorial argument)?
4. What are some real-world applications of this summation formula?
5. How do we extend the concept of mathematical induction to other structures like sets or graphs?

Tip: Induction is a powerful tool in mathematics because it allows you to prove statements about infinite sets by verifying a base case and a general inductive step.