Really hard to make it out, especially the upper limit of the sum, but I think it reads
[tex]\displaystyle\sum_{k=1}^n5k-42[/tex]
(not sure what [tex]n[/tex] is, but that's not very important)
Expand the sum as
[tex]\displaystyle5\sum_{k=1}^nk-42\sum_{k=1}^n1[/tex]
Recall that
[tex]\displaystyle\sum_{k=1}^n1=1+1+\cdots+1=n[/tex]
[tex]\displaystyle\sum_{k=1}^nk=1+2+\cdots+n=\dfrac{n(n+1)}2[/tex]
so that
[tex]\displaystyle\sum_{k=1}^n5k-42=\dfrac{5n(n+1)}2-42n[/tex]
Then just plug in whatever [tex]n[/tex] happens to be.
