Answer: [tex]\displaystyle \sum_{n=1}^{10} (3n-4)[/tex]
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Explanation:
This is an arithmetic sequence
[tex]a_1 = -1[/tex] is the first term
[tex]d = 3[/tex] is the common difference as we add 3 to each term to get the next term (eg: -1+3 = 2, or 2+3 = 5, or 5+3 = 8, etc)
let's find the nth term
[tex]a_n = a_1 + d(n-1)\\\\a_n = -1 + 3(n-1)\\\\a_n = -1 + 3n-3\\\\a_n = 3n-4\\[/tex]
Therefore, we are summing terms of the form 3n-4 from n = 1 to n = 10
With sigma notation, we would write
[tex]\displaystyle \sum_{n=1}^{10} (3n-4)[/tex]
This is a shorthand way to write out adding -1+2+5+8+11+...+26 where 26 is the tenth term. You would plug n = 10 into 3n-4 to get 26 as the tenth term.
