The image contains a mathematical induction proof for the formula:

$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$

Here’s a breakdown of the steps involved in this proof:

1. Base Case:

The proof starts with the base case $n = 1$.

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

So, the base case holds.

2. Inductive Hypothesis:

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

$1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2}$

3. Inductive Step:

The goal is to prove that the formula holds for $n = k + 1$, i.e.,

$1 + 2 + 3 + \dots + k + (k+1) = \frac{(k+1)((k+1) + 1)}{2} = \frac{(k+1)(k+2)}{2}$

Start with the left-hand side, which is the sum up to $k$, plus $k+1$:

$1 + 2 + 3 + \dots + k + (k+1) = \frac{k(k+1)}{2} + (k+1)$

Factor out $k+1$ on the right-hand side:

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

This is exactly the right-hand side of the formula for $n = k+1$.

4. Conclusion:

Since both the base case and the inductive step hold, by the principle of mathematical induction, the formula is true for all $n \geq 1$.

Would you like more details on any of the steps or another breakdown of a specific part?

Here are 5 follow-up questions related to the proof:

1. Why is mathematical induction a valid method of proof for this kind of problem?
2. Can this formula be derived using another method, such as summing techniques?
3. How would the proof change if the series started at 0 instead of 1?
4. Can you think of other series where mathematical induction can be applied similarly?
5. What is the geometrical interpretation of the formula $\frac{n(n+1)}{2}$?

Tip: When performing mathematical induction, make sure to carefully simplify expressions in the inductive step to avoid errors.