The problem in the image asks you to prove by mathematical induction that:

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

for all positive integers $n$.

Proof by Induction:

Step 1: Base Case ( $n = 1$ )

When $n = 1$, the left-hand side (LHS) is simply:

$1$

The right-hand side (RHS) is:

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

Thus, LHS = RHS when $n = 1$, so the base case holds.

Step 2: Inductive Hypothesis

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

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

This is called the inductive hypothesis.

Step 3: Inductive Step

Now, we need to prove that the statement holds for $n = k + 1$, i.e.,

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

Using the inductive hypothesis, we can write the left-hand side as:

$(1 + 2 + 3 + \cdots + k) + (k + 1)$

By the inductive hypothesis, this becomes:

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

Factor out $(k + 1)$:

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

This is exactly the right-hand side of the equation for $n = k + 1$. Therefore, the statement holds for $n = k + 1$.

Conclusion: By the principle of mathematical induction, the formula

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

is true for all positive integers $n$.

Would you like further clarification on any step? Here are some follow-up questions for deeper exploration:

1. How can we modify this induction process for different summation formulas?
2. What other types of problems can be solved using mathematical induction?
3. Why is the base case essential in mathematical induction?
4. How would the proof change if we were summing odd numbers instead of consecutive integers?
5. What is the principle behind strong induction, and how does it differ from regular induction?

Tip: When working with induction, always verify both the base case and the inductive step carefully. Missing out on one can invalidate the proof!