Answer:
[tex]\sum_{i=1}^{12}(3i-1) = 222[/tex]
Step-by-step explanation:
The given series is
[tex]\sum_{i=1}^{12}(3i-1)[/tex]
The first term is
[tex]f(1) = 3 \times 1 - 1 = 2[/tex]
The last term is
[tex]f(12) = 3 \times 12 - 1 = 35[/tex]
To find the number of terms in the sequence, solve,
[tex]3i - 1 = 35[/tex]
[tex]3i = 36 \\ i = 12[/tex]
The sum of the first i-terms of an arithmetic sequence is:
[tex]S_i = \frac{i}{2} (f(1) + f(2))[/tex]
We substitute the values to get;
[tex]S_ {12}= \frac{12}{2} (2+ 35)[/tex]
[tex]S_ {12}=6(37)[/tex]
[tex]S_ {12}= 222[/tex]
