completely simplify using PMI 7) Simplify the following: ∑ (𝑟 − 1)(𝑟 + 1) 𝑟=2𝑛 𝑟=1

We are asked to simplify the sum:

\[
\sum_{r=1}^{n} (r - 1)(r + 1)
\]

### Step 1: Expand the terms inside the summation
We start by expanding \((r - 1)(r + 1)\). Using the difference of squares:

\[
(r - 1)(r + 1) = r^2 - 1
\]

So the sum becomes:

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

### Step 2: Break the sum into two separate sums
Now, we can split this sum into two simpler sums:

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

The first sum is the sum of squares, and the second sum is simply a constant:

\[
\sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} 1
\]

### Step 3: Apply known summation formulas
We can now use the known formulas for these sums:

- The sum of squares:

\[
\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}
\]

- The sum of 1's:

\[
\sum_{r=1}^{n} 1 = n
\]

### Step 4: Combine the results
Substitute these formulas back into the equation:

\[
\frac{n(n+1)(2n+1)}{6} - n
\]

### Step 5: Simplify the result
We can factor \(n\) out of the second term to get a common denominator:

\[
\frac{n(n+1)(2n+1)}{6} - \frac{6n}{6}
\]

Now, combine the terms:

\[
\frac{n[(n+1)(2n+1) - 6]}{6}
\]

### Step 6: Expand and simplify
Let's expand \((n+1)(2n+1)\):

\[
(n+1)(2n+1) = 2n^2 + 3n + 1
\]

Now subtract 6:

\[
2n^2 + 3n + 1 - 6 = 2n^2 + 3n - 5
\]

So the expression becomes:

\[
\frac{n(2n^2 + 3n - 5)}{6}
\]

### Final Simplified Expression
Thus, the simplified sum is:

\[
\frac{n(2n^2 + 3n - 5)}{6}
\]

This is the final answer.

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Would you like more details or have any questions about this solution?

Here are five related questions to consider:
1. How can the sum of cubes be derived?
2. How does this result compare to a sum involving \(r^3 - 1\)?
3. Can the method of summation be applied to factorials?
4. What are other forms of summations that use quadratic terms?
5. How can generating functions simplify summations?

**Tip:** Always check the boundaries of your sum to ensure proper evaluation when splitting into smaller sums.

Explore how to simplify a summation involving the difference of squares and apply the sum of squares formula. This solution provides a detailed breakdown of the process, making it suitable for advanced high school students in Grades 11-12.

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