We are asked to prove the formula:

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

by mathematical induction.

Step 1: Base Case

Start by checking the base case when $n = 1$.

The left-hand side (LHS) is:

$1 = 1$

Now, calculate the right-hand side (RHS):

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

Since LHS = RHS, the base case holds true.

Step 2: Inductive Hypothesis

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

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

This is the inductive hypothesis.

Step 3: Inductive Step

We need to prove that the formula holds for $k + 1$, i.e.,

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

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

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

Now, simplify the right-hand side of the equation:

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

Combine the terms:

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

Factor out $(k+1)$:

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

This is exactly the formula for $n = k + 1$.

Step 4: Conclusion

Since the base case is true and the inductive step holds, by the principle of mathematical induction, the formula is valid for all natural numbers $n$.

Thus, we have proven that:

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

Follow-up:

• Would you like further details or clarification on any part of the proof?

Here are 5 related questions:

1. How does the principle of induction work in general?
2. Can this formula be derived using other methods like sums of sequences?
3. What is the closed-form expression for the sum of the first $n$ odd numbers?
4. How can we extend induction to prove formulas involving more complex summations?
5. How does this formula relate to the concept of arithmetic progressions?

Tip: When proving by induction, always clearly state both the base case and the inductive hypothesis. This step ensures that the proof logically flows to the desired conclusion.