•(〖n-1)〗^2- ((n-2)(n-1))/2 =(n-1)n/2      explain how answer came out and also factor out common factor

Let's solve and explain how the given expression:

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

Step by step:

### 1. Simplify the left-hand side expression
We begin by simplifying the left-hand side (LHS) step by step.

**Expression:**
\[
(n-1)^2 - \frac{(n-2)(n-1)}{2}
\]

#### First term: \((n-1)^2\)
Expand the square:
\[
(n-1)^2 = n^2 - 2n + 1
\]

#### Second term: \(\frac{(n-2)(n-1)}{2}\)
First, expand \((n-2)(n-1)\):
\[
(n-2)(n-1) = n^2 - 3n + 2
\]
Now divide this by 2:
\[
\frac{n^2 - 3n + 2}{2}
\]

Now, substitute the expanded forms back into the expression:
\[
(n^2 - 2n + 1) - \frac{(n^2 - 3n + 2)}{2}
\]

### 2. Combine the terms
To combine these two terms, let's rewrite the first part as a fraction with the same denominator (2):

\[
\frac{2(n^2 - 2n + 1)}{2} - \frac{n^2 - 3n + 2}{2}
\]

Now, subtract the numerators:
\[
\frac{2(n^2 - 2n + 1) - (n^2 - 3n + 2)}{2}
\]

Expand and simplify:
\[
2(n^2 - 2n + 1) = 2n^2 - 4n + 2
\]

So:
\[
\frac{(2n^2 - 4n + 2) - (n^2 - 3n + 2)}{2}
\]
\[
= \frac{2n^2 - 4n + 2 - n^2 + 3n - 2}{2}
\]
\[
= \frac{(2n^2 - n^2) + (-4n + 3n) + (2 - 2)}{2}
\]
\[
= \frac{n^2 - n}{2}
\]

So, the simplified left-hand side is:
\[
\frac{n(n-1)}{2}
\]

### 3. Right-hand side
The right-hand side of the equation is already given as:
\[
\frac{n(n-1)}{2}
\]

### 4. Conclusion
Since the left-hand side equals the right-hand side, the equation is correct:

\[
\boxed{\frac{n(n-1)}{2} = \frac{(n-1)n}{2}}
\]

### Factor out the common factor:
We see that \(\frac{n(n-1)}{2}\) already has \(n(n-1)\) as a common factor in the numerator, so this is the factored form of the final answer.

### Summary:
- Expanded and simplified each term.
- Combined them under a common denominator.
- Simplified the final expression, showing it matches the right-hand side.

---

Would you like more details or further clarification? 

Here are 5 relative questions for deeper understanding:
1. How do you expand binomials like \((n-1)^2\)?
2. What is the process of combining fractions with different terms?
3. How can common factors be identified in polynomial expressions?
4. What steps are involved in factoring quadratic expressions?
5. How does factoring simplify solving equations?

**Tip:** When dealing with fractions, always try to get a common denominator to simplify complex expressions easily.

•(〖n-1)〗^2 - ((n-2)(n-1))/2 =(n-1)n/2 explain how answer came out and also factor out common factor

Explore a step-by-step solution for simplifying the expression (n-1)^2 - ((n-2)(n-1))/2 and factoring out the common terms to arrive at the final equation (n-1)n/2. This solution covers key algebraic concepts such as polynomial expansion, common denominators, and factoring techniques. Suitable for students in Grades 9-12.

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