Answer:
[tex]\displaystyle \boxed{\sum_{n=1}^{19} (n^2 - 6n + 9) - \sum_{n=1}^{9} (n^2 - 6n + 9) = \sum_{n=10}^{19} (n-3)^2}.[/tex]
Step-by-step explanation:
We can start by noticing that
[tex]n^2 - 6n + 9 = n^2 - 2 \times 3 \times n + 3^2 = (n-3)^2,[/tex]
so we get:
[tex]\displaystyle \sum_{n=1}^{19} (n^2 - 6n + 9) - \sum_{n=1}^{9} (n^2 - 6n + 9) = \sum_{n=1}^{19} (n-3)^2 - \sum_{n=1}^{9} (n-3)^2.[/tex]
Let [tex]a_n=(n-3)^2[/tex]. We can now write the expression as:
[tex]\displaystyle\sum_{n=1}^{19} a_n - \sum_{n=1}^{9} a_n = (a_1 + a_2 + \dots + a_{19}) - (a_1 + a_2 + \dots + a_9).[/tex]
Since the first 9 terms cancel, we get:
[tex]\displaystyle a_{10} + a_{11} + \dots + a_{19} = \sum_{n=10}^{19}a_n.[/tex]
So we finally get:
[tex]\displaystyle \boxed{\sum_{n=1}^{19} (n^2 - 6n + 9) - \sum_{n=1}^{9} (n^2 - 6n + 9) = \sum_{n=10}^{19} (n-3)^2}.[/tex]
