Formula for consecutively subtracting a percentage n times

I'm looking for a formula to subtract a percentage (e.g. $$10\%$$) from a number (e.g. $$100$$) a certain number of times (e.g. $$4$$) in which case:

$$x=100$$

$$p=0,1$$ (i.e. $$10\%$$)

$$n=4$$

Doing it manually goes like this:

1. $$y_1 = 100 - 10\% = 90$$

2. $$y_2 = 90 - 10\% = 81$$

3. $$y_3 = 81 - 10\% = 72.9$$

4. $$y_4 = 72.9 - 10\% = 65.61$$

The formulas for these steps I've managed to figure out are:

1. $$y_1 = x - 1\cdot px$$
2. $$y_2 = x - 2\cdot px + 1\cdot p^2x$$
3. $$y_3 = x - 3\cdot px + 3\cdot p^2x - 1\cdot 0.1^3x$$
4. $$y_4 = x - 4\cdot px + 6\cdot p^2x - 4\cdot0.1^3x + 0.1^4x$$

The formula 'grows' with each iteration.

Is there a single universal formula that would use $$x$$, $$p$$ to find $$y_n$$ for any $$n$$?

I suppose in function notation the question would look like $$y = f(x,p,n)$$.

• It's very useful to make the mental connection that "subtracting" 10 per cent is actually multiplying by 0.9. So you're repeatedly multiplying by 0.9. Jun 3, 2021 at 9:16

Note that instead of subtracting $$0.1\cdot y_{n-1}$$ from $$y_{n-1}$$, you can simple multiply $$(1-0.1)\cdot y_{n-1}$$. The factor $$(1-0.1)$$ doesn't depend on $$n$$, so the formula becomes much easier: since we multiply by $$(1-0.1)$$ at each iteration, we effectively multiply by $$(1-0.1)^n$$ after $$n$$ iterations.
$$y_n=(1-p)^n x.$$