# Formula for series (or sequence?) of $n$ values where last in series is $x \times n_1$ and total is $y$.

I am a programmer, not a mathematician and I'm trying to solve a problem for scoring in my game. Please excuse my weak maths and lack of mathematical syntax, and sorry if this question has been asked before.

Let's say we have a game level which consists of 4 stages. As the players moves from stage to stage, the required score for each stage increases. Let's say the overall score for the level is 1000 and I want the last stage to require 4 times more score.

Let these be the variables:

• $$y$$ is level total required score.
• $$n_1, n_2, n_3, n_4$$ are 4 stages in level.
• $$x$$ is the score requirement multiplier for the last stage ($$d$$).

So to represent as an equation:

$$n_1 + n_2 + n_3 + n_4 = y$$ where $$n_4 = x \times n_1$$

therefore based on the values in my example the equation would be:

$$n_1 + n_2 + n_3 + 4n_1 = 1000$$

How do I solve for $$n_1$$ , $$n_2$$ and $$n_3$$ such that there is a progression in the score requirement from stage to stage?

For some reason this formula works, substituting for the variables:

$$n_1 + \frac {4n_1}{3} + \frac {4n_1}{2} + 4n_1 = 1000$$

I able to solve for $$n_1$$ and I get $$n_1 = 120$$.

This answer checks out:

$$120 + \frac {4 \times 120}{3} + \frac {4 \times 120}{2} + (4 \times 120) = 1000$$

$$120 + 160 + 240 + 480 = 1000$$

But I don't understand how that works. Where do the denominators for the fractions $$\frac 43$$ and $$\frac 42$$ come from?

Anyone that could shed some light on how to solve this, your time would be appreciated :)

• It looks like a geometric sum. You're starting with 120. Then you add 40, and then $2\times 40$ and then $2^2 \times 40 ...$ So you're basically calculating $$120 + 40 \sum_{k=0}^n 2^k$$ Nov 21, 2018 at 9:18
• Thanks @MattiP. I wish I could read that formula... :) I think I understand the gist of it.
– jnt
Nov 21, 2018 at 9:36

This question is a little difficult to answer mathematically, since $$a+b+c+xa=1000$$ has many possible solutions. So let's refine it a little bit. Let's add a requirement that each stage costs $$r$$ times as much as the previous stage. We can then rewrite the equation as $$a+ar+ar^2+ar^3=1000.$$ At this point, you can just pick your favorite value of $$r$$ and solve for $$a$$. If you want each stage to cost $$2$$ times as much as the previous stage then $$a+2a+4a+8a=1000$$ $$15a=1000$$ $$a=\frac{200}3 \approx 67.$$ This gives the value of the first stage, and then $$2a, 4a,$$ and $$8a$$ are the values of the stages afterward.