Solve for stopping point of sum

Question

I am currently in 8th grade and in my Algebra class we are currently covering exponential growth, e.g. bacteria splitting, fruit fly growth, etc. Anyway, one exercise we did was on zombies. The problem started with 5 "sleeper cell" zombies, who each day could "turn" three humans. The equation looked like this: $5\times 3^x = h$

$h$ = humans infected each day.

While this was interesting, I wanted to see how many total infected humans there were including the original 5 "sleeper cell" zombies. I had a scientific calculator on hand so I calculated the sum of all of this with $\displaystyle \sum_{d=0}^{19}5(3)^d$ where $d = $ days passed.

This ended up being 8,716,961,000, a little more than the total Earth's population. Therefore $d>18$ but $d<19$ in order for the sum to $\approx$ 7,000,000,000 (Earth's population).

My question is, is it possible to solve for the stopping point of the sum/upper value of the sum? Please keep in mind that although I am in AP 8th grade math, I'm no mathematician, so please keep the explanations within the realm of my comprehension.

Answer 1

This question is nice!

Okay let's say the sum for the first $n$ terms is $s_n$ (just forget about the 5 first):

$$ s_n = \sum_{i=0}^n 3^i $$ Then we know that:

$$ 3\cdot s_n =\sum_{i=0}^n3^{i+1} =3^1+3^2+3^3+\dots+3^{n+1} \\ s_n=\sum_{i=0}^n3^i=3^0+3^1+3^2+\dots+3^n$$

So let's substract those sums:

$$ s_n - 3\cdot s_n = \sum_{i=0}^n3^{i}-3^{i+1}=3^0+\underbrace{(-3^1+3^1)}_{=0}+(-3^2+3^2)+\cdots-3^{n+1} = 3^0-3^{n+1}$$

So then we know:

$$ s_n-3\cdot s_n = s_n(1-3) = 1-3^{n+1}$$ So let's just try to get the $n$: $$ 3^{n+1} = 1+2s_n$$

Now we have to take the $\log_3$:

$$ n+1 = \log_3\left(1+2s_n\right) = \frac{\log(1+2s_n)}{\log(3)}$$

So wait we forgot the $5$ right, so your stopping value $S$ is five times bigger: $$ S = 5\cdot s_n$$

Giving you the formula:

$$ n = \frac{\log\left(1+\frac{2}{5}S\right)}{\log(3)}-1 $$

Answer 2

I think there's something not quite right about the equation $h=5\times3^x$.

Look at it this way. You start day 1 with $5$ zombies. Each of them "turns" three humans that day. So at the end of the day you have the original $5$ zombies plus $15$ new zombies, for a total of $20$. Now on day 2, those $20$ zombies each turn three humans, so at that end of the second day you have the $20$ zombies you started with plus $60$ new zombies, for a total of $80$. At the end of day 3 you'll have $80+240=320$ zombies, and so forth.

The point is, zombies don't die (unless you shoot them in the head, of course). So it seems to me that the basic equation is

$$Z(x)=\text{ the number of zombies at the end of day $x$ }=5\times4^x$$

If you want to know the number of humans turned on day $x$, you might write that as

$$h(x)=Z(x)-Z(x-1)=5(4^x-4^{x-1})=15\times4^{x-1}$$

But the total number of humans turned by the end of day $x$ is simply

$$Z(x)-5=5(4^x-1)$$

If you want to know when this number exceeds the population of the Earth, note that $4^5=2^{10}\approx1000$, hence $4^{15}=2^{30}\approx1$ billion. Multiplying by $5$ and comparing to the approximate human population of $7$ billion, we see that the zombies run out of people to turn on the 16th day.

Answer 3

This is a geometric series The general equation, if $r$ is the ratio between the terms ($3$ for you) is $\sum_{i=1}^d r^i=\frac {r^{d+1}-1}{r-1}$ You can distribute out the $5$. You can use logaritms to solve for $d$