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.