The coefficient on [tex]x[/tex] starts at 14 and decreases by 2 until it reaches 0. Similarly, the coefficient on [tex]y[/tex] starts at 0 and increases by 1 until it reaches 7.
So the general [tex]n[/tex]-th term of the sum might by
[tex](14-2(n-1))x+(n-1)y[/tex]
where [tex]n[/tex] starts at 1. Because the last term in the sum is [tex]7y[/tex], we deduce that there are [tex]n-1=7\implies n=8[/tex] terms in the entire sum, so we have
[tex]\displaystyle\sum_{n=1}^8\bigg((14-2(n-1))x+(n-1)y\bigg)=\sum_{n=1}^8\bigg((16-2n)x+(n-1)y\bigg)[/tex]
