The given problem involves proving the property $P_n$ for $n \in \mathbb{N}$:

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

This is a classical result, and we can prove it by mathematical induction.

Step 1: Base Case $n = 1$

For $n = 1$, the left-hand side of the equation is simply: $1 = 1$ The right-hand side is: $\frac{1(1+1)}{2} = \frac{1 \times 2}{2} = 1$ Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Assume that the property holds for some $n = k$, i.e., $1 + 2 + \cdots + k = \frac{k(k+1)}{2}$ This is our inductive hypothesis.

Step 3: Inductive Step

We now need to prove that if the property holds for $n = k$, then it also holds for $n = k+1$, i.e.,

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

Using the inductive hypothesis, the left-hand side becomes: $\left(1 + 2 + \cdots + k\right) + (k+1) = \frac{k(k+1)}{2} + (k+1)$ Now, factor out $(k+1)$: $\frac{k(k+1)}{2} + \frac{2(k+1)}{2} = \frac{k(k+1) + 2(k+1)}{2}$ $= \frac{(k+1)(k+2)}{2}$ This matches the right-hand side of the equation for $n = k+1$. Hence, the property holds for $n = k+1$.

Conclusion

By the principle of mathematical induction, the property $P_n : 1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ is true for all $n \in \mathbb{N}$.

Would you like more details or have any questions about this?

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Here are five related questions to deepen your understanding:

1. How does mathematical induction work, and why is it a valid proof technique?
2. What are some common patterns in sums, like the sum of squares or cubes?
3. How can we derive a closed-form formula for the sum of an arithmetic series?
4. What is the connection between this sum and the area of a triangle in geometry?
5. Can you prove this formula using other methods, such as a combinatorial argument?

Tip: Always verify the base case carefully when using induction—it ensures the foundation of the proof!