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?

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.