We are asked to prove by induction that:

$\sum_{i=0}^{k} i = \frac{k(k+1)}{2} \quad \text{for all} \quad k \in \mathbb{N}.$

Step 1: Base Case

We first check the base case, $k = 0$.

The left-hand side of the equation is the sum:

$\sum_{i=0}^{0} i = 0.$

The right-hand side is:

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

Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Assume the statement is true for $k = n$, i.e.,

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

Step 3: Inductive Step

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

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

Start with the left-hand side:

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

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

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

Now, factor out $(n + 1)$:

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

Simplify the numerator:

$n(n+1) + 2(n+1) = (n+1)(n + 2).$

Thus, we have:

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

This matches the right-hand side of the statement for $k = n + 1$. Therefore, the inductive step is complete.

Conclusion

By the principle of mathematical induction, we have shown that:

$\sum_{i=0}^{k} i = \frac{k(k+1)}{2} \quad \text{for all} \quad k \in \mathbb{N}.$

Would you like further details on any part of the proof or have any questions?

Here are five related questions to expand your understanding:

1. Can you use a similar method to prove the formula for the sum of squares?
2. How would you modify the approach if the sum started at 1 instead of 0?
3. What other mathematical statements can be proven using induction?
4. How does the inductive hypothesis work in more complex summations or sequences?
5. Can this summation formula be derived using another method, such as direct counting or combinatorics?

Tip: When proving a statement by induction, always ensure your base case is clearly checked before moving on to the inductive step.