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The given equation is:

\[
\sum_{k=1}^{n-1} (-1)^{k-1} k^2 = (-1)^n \cdot \frac{(n-1)n}{2}
\]
where \( n \geq 2 \).

Let's break this down step by step.

### Left-hand side: Summation
The summation term is:

\[
\sum_{k=1}^{n-1} (-1)^{k-1} k^2
\]

This is an alternating sum of the squares of integers from \( k = 1 \) to \( k = n-1 \), where the sign alternates based on \( (-1)^{k-1} \).

### Right-hand side:
The right-hand side is:

\[
(-1)^n \cdot \frac{(n-1)n}{2}
\]

This is a product of \( (-1)^n \) and the arithmetic series sum of the first \( n-1 \) integers, which gives \( \frac{(n-1)n}{2} \).

### General approach:
We need to verify the equality holds for general values of \( n \geq 2 \). Let's test the formula for small values of \( n \) to verify the pattern.

#### Case 1: \( n = 2 \)

- Left-hand side:

\[
\sum_{k=1}^{2-1} (-1)^{k-1} k^2 = (-1)^0 \cdot 1^2 = 1
\]

- Right-hand side:

\[
(-1)^2 \cdot \frac{(2-1) \cdot 2}{2} = 1
\]

Both sides are equal.

#### Case 2: \( n = 3 \)

- Left-hand side:

\[
\sum_{k=1}^{3-1} (-1)^{k-1} k^2 = (-1)^0 \cdot 1^2 + (-1)^1 \cdot 2^2 = 1 - 4 = -3
\]

- Right-hand side:

\[
(-1)^3 \cdot \frac{(3-1) \cdot 3}{2} = (-1) \cdot \frac{2 \cdot 3}{2} = -3
\]

Again, both sides are equal.

### Conclusion:
The equation holds for these values of \( n \), and the structure suggests it will hold for any \( n \geq 2 \). This alternating sum of squares formula is valid for \( n \geq 2 \).

Would you like further details on how to prove this more rigorously for any \( n \), or any other related concepts?

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Here are 5 related questions you could explore:
1. How does the alternating sign impact the summation process in this problem?
2. Can we find a closed-form solution for the sum of squares of integers without alternating signs?
3. What is the significance of \( \frac{(n-1)n}{2} \) in arithmetic progressions?
4. How does this formula generalize for higher powers, such as \( k^3 \)?
5. What role do alternating series play in calculus and series summations?

**Tip:** When verifying identities, always test small values of variables first to get a sense of how the formula behaves.

Solve the equation: \(\sum_{k=1}^{n-1} (-1)^{k-1} k^2 = (-1)^n \cdot \frac{(n-1)n}{2}\), where \(n \geq 2\).

This problem involves solving an alternating series summation with squared terms, where the sum alternates based on the powers of -1. The solution involves the sum of squares and an arithmetic progression formula for \(n \geq 2\). Suitable for advanced high school students.

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