Let's prove the given statement by mathematical induction.

Statement:

We need to prove that for any positive integer $n$, the following equation holds:

$S(n) = -1^2 + 2^2 - 3^2 + 4^2 - \ldots + (-1)^n \cdot n^2 = (-1)^n \cdot \frac{n(n+1)}{2}$

Step 1: Base Case

Let's check the base case $n = 1$:

For $n = 1$:

$S(1) = (-1)^1 \cdot 1^2 = -1$

On the other side:

$(-1)^1 \cdot \frac{1(1+1)}{2} = -1 \cdot \frac{1 \cdot 2}{2} = -1 \cdot 1 = -1$

The base case holds true since both sides are equal.

Step 2: Inductive Hypothesis

Assume the statement is true for $n = k$, where $k$ is a positive integer:

$S(k) = -1^2 + 2^2 - 3^2 + 4^2 - \ldots + (-1)^k \cdot k^2 = (-1)^k \cdot \frac{k(k+1)}{2}$

Step 3: Inductive Step

We need to show that the statement holds true for $n = k + 1$:

Consider the sum $S(k+1)$:

$S(k+1) = S(k) + (-1)^{k+1} \cdot (k+1)^2$

Using the inductive hypothesis:

$S(k+1) = (-1)^k \cdot \frac{k(k+1)}{2} + (-1)^{k+1} \cdot (k+1)^2$

Factor out $(-1)^{k+1}$ from the right-hand side:

$S(k+1) = (-1)^{k+1} \left[ -\frac{k(k+1)}{2} + (k+1)^2 \right]$

Simplify the expression inside the brackets:

$S(k+1) = (-1)^{k+1} \left[ \frac{-k(k+1) + 2(k+1)^2}{2} \right]$

Factor out $(k+1)$ from both terms in the numerator:

$S(k+1) = (-1)^{k+1} \left[ \frac{(k+1)(-k + 2(k+1))}{2} \right]$

Simplify further:

$S(k+1) = (-1)^{k+1} \left[ \frac{(k+1)(-k + 2k + 2)}{2} \right]$

$S(k+1) = (-1)^{k+1} \left[ \frac{(k+1)(k + 2)}{2} \right]$

Finally, we see that:

$S(k+1) = (-1)^{k+1} \cdot \frac{(k+1)((k+1) + 1)}{2}$

This is exactly the form we wanted to prove, so the inductive step holds.

Conclusion:

By the principle of mathematical induction, the given statement is true for all positive integers $n$.

Would you like further clarification on any part of this proof or have any other questions?

Here are 5 relative questions you might find useful:

1. How can induction be applied to prove inequalities?
2. What is the significance of the base case in mathematical induction?
3. How do you modify the inductive hypothesis if you are proving a statement about a sequence?
4. Can you use induction to prove statements about divisibility?
5. How do you decide the appropriate base case for induction?

Tip: When performing the inductive step, always ensure that your algebraic manipulations are correct, as a small mistake can lead to incorrect conclusions.