We are given the equation:

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

Let's solve it step by step.

Step 1: Expand the terms

Start by expanding each term on the left-hand side.

1. Expand $(n-1)^2$:

$(n-1)^2 = n^2 - 2n + 1$

2. Expand $\frac{(n-2)(n-1)}{2}$:

$(n-2)(n-1) = n^2 - 3n + 2$ Thus: $\frac{(n^2 - 3n + 2)}{2} = \frac{n^2}{2} - \frac{3n}{2} + 1$

Step 2: Substitute the expanded terms

Substitute both expansions into the original equation:

$(n^2 - 2n + 1) - \left( \frac{n^2}{2} - \frac{3n}{2} + 1 \right) = \frac{(n-1)n}{2}$

Step 3: Simplify the left-hand side

Now, combine like terms on the left-hand side:

$n^2 - 2n + 1 - \left( \frac{n^2}{2} - \frac{3n}{2} + 1 \right)$

Distribute the negative sign:

$n^2 - 2n + 1 - \frac{n^2}{2} + \frac{3n}{2} - 1$

Now, combine like terms:

$n^2 - \frac{n^2}{2} = \frac{n^2}{2}, \quad -2n + \frac{3n}{2} = -\frac{n}{2}, \quad 1 - 1 = 0$

So the left-hand side simplifies to:

$\frac{n^2}{2} - \frac{n}{2}$

Step 4: Compare both sides

Now, the equation becomes:

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

Expand the right-hand side:

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

Thus, we have:

$\frac{n^2}{2} - \frac{n}{2} = \frac{n^2}{2} - \frac{n}{2}$

Both sides are equal, so the equation is true.

Conclusion:

The equation holds true for all values of $n$.

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Would you like further clarification on any part of the solution?

Here are five related questions to explore:

1. How can you solve quadratic equations with fractions?
2. How would you approach simplifying more complex algebraic fractions?
3. What is the process for expanding binomials like $(n-1)^2$?
4. What happens when you encounter an equation that seems unsolvable? How do you analyze it?
5. How does solving linear equations compare to solving quadratic ones?

Tip: Always expand and simplify carefully when working with equations involving squares and fractions to avoid small errors.