Is there another mathematical way to approach this problem? 
$$f(x)-f(x-1) = x-5$$
$$f(16) = 74$$
Compute $f(1)$.

So, this is basically a linear function. I, however, calculated it without taking the easier way to approach this problem.
If
$$f(16) - f(15) = 11$$
Then
$$74-f(15) = 11 \implies f(15)=63$$
This will take so long as it seems. Since we're given a linear function, how would we compute it in an other mathematical way? Furthermore, is there a general rule for what you advise?

First, note that $f$ isn't linear at all (if it were linear, then $f(x)-f(x-1)$ would be constant. However, that difference does suggest something quadratic: its second difference will be constant. Thus, we'll try to find $a$, $b$, and $c$ such that $f(x) = ax^2 + bx + c$ satisfies this. For this purpose, we garner some equations relating $a$, $b$, and $c$: 
First, $f(16) = 74$ tells us that $256a + 16b + c = 74$. Second, $f(x)-f(x-1) = x - 1$ gives us $a(x^2 - (x-1)^2) + b(x - (x-1)) = x - 1$. Simplifying that gives us $a(2x-1)+b=x-1$, for all $x$. Taking $x = 0$ gives us $b - a = -1$, $x = 1$ give $a + b = 0$, so $b = -a$, and the $x = 0$ case gives $2a = 1$, so $a = \frac{1}{2}$, and $b = \frac{-1}{2}$. Finally, $f(16)=74$ gives us $120 + c = 74$, so $c = -46$. Thus, $f(x) = \frac{1}{2}x^2 - \frac{1}{2}x - 46$. 
You can quickly go back and check that this satisfies all relevant conditions, if you like (it does), and therefore $f(1) = \frac{1}{2}(1^2) - \frac{1}{2}(1)-46 = -46$. 
