C. 392
Given the series:
[tex]\sum_{a=1}^{7} (3a^{2} - 4)[/tex]
We are told to evaluate it. This means that we substitute the value of a into the equation and then obtain the product and add it with the following products until we stop at the top value for a.
So, it would become:
[tex]\sum_{a=1}^{7} (3a^{2} - 4) = \\(3\times1^2 - 4) + (3\times2^2 - 4) + (3\times3^2 - 4) + (3\times4 ^2 - 4) + (3\times5^2 - 4) + (3\times6 ^2 - 4) + (3\times7^2 - 4)\\\\[/tex]
Solve.
[tex](3\times1^2 - 4) + (3\times2^2 - 4) + (3\times3^2 - 4) + (3\times4 ^2 - 4) + (3\times5^2 - 4) + (3\times6 ^2 - 4) + (3\times7^2 - 4)\\\\ = -1 + 8 + 23 + 44 + 71 + 104 + 143 \\= 392[/tex]
Thus, the value of the series is 392.
