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The given sum is:

\[
\sum_{i=1}^{28} (i - 1)^2
\]

To evaluate this summation, let's break it down.

1. The expression inside the summation is \((i - 1)^2\), which we need to sum from \(i = 1\) to \(i = 28\).

2. To simplify, notice that when \(i = 1\), \((i - 1)^2 = 0^2 = 0\), and for other values of \(i\), we sum the squares of integers up to \(27\) (since \(i-1\) ranges from 0 to 27).

So, this sum becomes:

\[
\sum_{i=1}^{28} (i - 1)^2 = \sum_{i=0}^{27} i^2
\]

This is a standard sum of squares, and the formula for the sum of squares of the first \(n\) integers is:

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

In this case, \(n = 27\). Plugging this into the formula:

\[
\sum_{i=1}^{27} i^2 = \frac{27(27+1)(2 \times 27 + 1)}{6} = \frac{27(28)(55)}{6}
\]

Now, simplify:

\[
\frac{27 \times 28 \times 55}{6} = 6930
\]

Thus, the value of the sum is:

\[
\sum_{i=1}^{28} (i - 1)^2 = 6930
\]

Would you like further clarification on any of the steps?

Here are some follow-up questions you might explore:
1. What is the general formula for summing cubes of integers?
2. How would the sum change if we summed \((i-1)^3\) instead of \((i-1)^2\)?
3. Can you derive the formula for the sum of squares of integers from scratch?
4. How does this method extend to summing powers higher than squares?
5. What happens if you change the upper bound (28) to another number?

**Tip**: Always double-check your final summation using graphing or summation tools to verify complex calculations!

Evaluate the sum \( \sum_{i=1}^{28} (i - 1)^2 \). Use the properties of summation and the Summation Formulas Theorem to find the result.

This guide demonstrates how to evaluate the sum \( \sum_{i=1}^{28} (i - 1)^2 \) using the properties of summation and the Sum of Squares formula. Learn step-by-step how to apply formulas for sequences and series, suitable for high school students.

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