Distribute the sum:
[tex]\displaystyle\sum_{i=1}^{24}(3i-2)=3\sum_{i=1}^{24}i-2\sum_{i=1}^{24}1[/tex]
Use the following formulas:
[tex]\displaystyle\sum_{i=1}^n1=n[/tex]
[tex]\displaystyle\sum_{i=1}^ni=\dfrac{n(n+1)}2[/tex]
[tex]\implies\displaystyle\sum_{i=1}^{24}(3i-2)=3\cdot\frac{24\cdot25}2-2\cdot24=\boxed{852}[/tex]
In case you don't know where those formulas came from:
The first one is obvious; you're just adding n copies of 1, so 1 + 1 + ... + 1 = n.
The second can be proved in this way: let S be the sum 1 + 2 + 3 + ... + n. Rearrange it as S = n + (n - 1) + (n - 2) + ... + 1. Then 2S = (n + 1) + (n + 1) + (n + 1) + ... + (n + 1), or n copies of n + 1. So 2S = n(n + 1). Divide both sides by 2 and we're done.
