Which of the following is equivalent to \( \sum_{n=1}^{19}\left(n^{2}-6 n+9\right)-\sum_{n=1}^{9}\left(n^{2}-6 n+9\right) \) ? \( \sum_{n=10}^{19}(n-3)^{2} \) \( \sum_{n=8}^{19}(n-3)^{2} \) \( \sum_{n=10}^{19}(n+3)^{2} \) \( \sum_{n=8}^{19}(n+3)^{2} \)

Let's break down the expression step-by-step to find its equivalent. The expression is: \[ \sum_{n=1}^{19}(n^{2}-6n+9) - \sum_{n=1}^{9}(n^{2}-6n+9) \] Notice that \( n^{2} - 6n + 9 = (n-3)^2 \). Thus, we can rewrite the sums as follows: \[ \sum_{n=1}^{19} (n-3)^2 - \sum_{n=1}^{9} (n-3)^2 \] This means we are calculating the difference between the two sums. However, we can simplify it directly by noting the behavior of the sums. The first sum, \(\sum_{n=1}^{19} (n-3)^2\), represents the sum of squares from \(n=1\) to \(n=19\), while the second sum represents the squares from \(n=1\) to \(n=9\). Thus, we only need the main series from 10 to 19: \[ \sum_{n=10}^{19} (n-3)^2 \] Now, substituting \(n-3\) in the range \(n=10\) to \(19\) gives us the terms that fall as follows: - For \(n=10\), \(n-3 = 7\). - For \(n=19\), \(n-3 = 16\). This sums the squares of the integers 7 through 16, which confirms our expression equals: \[ \sum_{n=10}^{19} (n-3)^2 \] Thus, the correct answer is: \[ \sum_{n=10}^{19}(n-3)^{2} \] This matches the first option provided: \(\sum_{n=10}^{19}(n-3)^{2}\).