Math Problem Statement

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

Solution

Let's solve and explain how the given expression:

(n1)2(n2)(n1)2=(n1)n2\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: (n1)2(n2)(n1)2(n-1)^2 - \frac{(n-2)(n-1)}{2}

First term: (n1)2(n-1)^2

Expand the square: (n1)2=n22n+1(n-1)^2 = n^2 - 2n + 1

Second term: (n2)(n1)2\frac{(n-2)(n-1)}{2}

First, expand (n2)(n1)(n-2)(n-1): (n2)(n1)=n23n+2(n-2)(n-1) = n^2 - 3n + 2 Now divide this by 2: n23n+22\frac{n^2 - 3n + 2}{2}

Now, substitute the expanded forms back into the expression: (n22n+1)(n23n+2)2(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):

2(n22n+1)2n23n+22\frac{2(n^2 - 2n + 1)}{2} - \frac{n^2 - 3n + 2}{2}

Now, subtract the numerators: 2(n22n+1)(n23n+2)2\frac{2(n^2 - 2n + 1) - (n^2 - 3n + 2)}{2}

Expand and simplify: 2(n22n+1)=2n24n+22(n^2 - 2n + 1) = 2n^2 - 4n + 2

So: (2n24n+2)(n23n+2)2\frac{(2n^2 - 4n + 2) - (n^2 - 3n + 2)}{2} =2n24n+2n2+3n22= \frac{2n^2 - 4n + 2 - n^2 + 3n - 2}{2} =(2n2n2)+(4n+3n)+(22)2= \frac{(2n^2 - n^2) + (-4n + 3n) + (2 - 2)}{2} =n2n2= \frac{n^2 - n}{2}

So, the simplified left-hand side is: n(n1)2\frac{n(n-1)}{2}

3. Right-hand side

The right-hand side of the equation is already given as: n(n1)2\frac{n(n-1)}{2}

4. Conclusion

Since the left-hand side equals the right-hand side, the equation is correct:

n(n1)2=(n1)n2\boxed{\frac{n(n-1)}{2} = \frac{(n-1)n}{2}}

Factor out the common factor:

We see that n(n1)2\frac{n(n-1)}{2} already has n(n1)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 (n1)2(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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomial Expansion
Simplification of Expressions

Formulas

(a-b)^2 = a^2 - 2ab + b^2
Multiplying binomials
Factoring out common terms

Theorems

Binomial Expansion Theorem
Distributive Property

Suitable Grade Level

Grades 9-12