Respuesta :
The series is an arithmetic series with first term -10 and common difference (-2-(-10))=8. The general term a_n can be written as
a_n = -10 + 8(n-1)
or
a_n = -18+8n
The last term corresponds to
110 = -18+8n
128 = 8n
16 = n
So, the summation can be written as
[tex]S_{16}=\sum\limits_{n=1}^{16}{(-18+8n)}[/tex]
a_n = -10 + 8(n-1)
or
a_n = -18+8n
The last term corresponds to
110 = -18+8n
128 = 8n
16 = n
So, the summation can be written as
[tex]S_{16}=\sum\limits_{n=1}^{16}{(-18+8n)}[/tex]
The answer is summation of the quantity negative ten plus eight n from n equals zero to fifteen