Respuesta :

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]