ANSWER
[tex] \sum _{n = 1} ^{18} (6n - 15)[/tex]
EXPLANATION
The given series
[tex] - 9 - 3 + 3 + 9 + ... + 81[/tex]
This is an arithmetic series with a common difference of
[tex]d = - 3 - - 9 = 6[/tex]
The first term of the series is:
[tex]a_1 = - 9[/tex]
The general term is given by:
[tex]a_n =a_1 + d(n - 1)[/tex]
[tex]a_n = - 9+ 6(n - 1)[/tex]
[tex]a_n = - 9+ 6n - 6[/tex]
[tex]a_n =6n - 15[/tex]
The last term is 81
We can use this to determine the number of terms in the series.
[tex]81=6n - 15[/tex]
[tex]81 + 15=6n [/tex]
[tex]6n = 96[/tex]
[tex]n = \frac{96}{6} = 16[/tex]
The summation notation is:
[tex] \sum _{n = 1} ^{18} (6n - 15)[/tex]