Vtyjkyuk

"The figure below shows the quantity of the drug atenolol in the blood as a function of time, with the first dose at time $t = 0$.
Suppose atenolol is taken in $75$ mg doses once a day to lower blood pressure. If the half-life of the atenolol in the blood is $6.4$ hours, the percentage of atenolol still present at the end of a $24$ hours period is approximately $7$ %."

I can't post the graph here, but $Q_n$ represents the amounts added, $P_n$ represents the remainder of previous doses still in the system. So $Q_0 = 75$, but $P_0=0$.

The first three terms for $Q_n$ are: $Q_0=75, Q_1=75+(75cdot 0.07), Q_2=75+(75cdot 0.07)+(75cdot 0.07^2)$ However I can't seem to get the closed form for this correct - I don't really understand these, but we were given two formulas, one for finite, one for infinite series, and I've tried both but neither seems to work. My teacher gave me a hint, "note that $n$ starts from zero, so $Q_n$ is in fact the sum of the first $n+1$ terms." But....I don't know what that means for trying to come up with a closed form.

The first three terms for $P_n$ are: $P_1 = (75cdot 0.07), P_2=(75cdot 0.07)+(75cdot 0.07^2), P_3=(75cdot 0.07)+(75cdot 0.07^2)+(75cdot 0.07^3)$

In the book the closest example has the form $fracacdot (1-x^n) 1-x$ where $a$ would equal $75$ and $x$ would equal $0.07$. I've tried everything I can think of here but I don't know what to do next.

Hint: it seems that $Q_n = 75frac0.07^n+1-10.07-1$, while $P_n = 75left(frac0.07^n+1-10.07-1-1right)$

When trying to use the general equations for the closed-formula evaluation
of geometric series, it's a good idea always to write down both sides of the equation.
The formula you wrote, $fraca(1-x^n) 1-x,$
is only one side of the equation, and it does not really tell you anything.
The full equation you should be looking at is
$$
a + ax + ax^2 + ax^3 + cdots + ax^n-1 = fraca(1-x^n) 1-x.
$$

Note that some people prefer the terms on the left to go all the
way up to $ax^n$ and so they have $x^n+1$ instead of $x^n$ on the right.
The two formulas are equivalent but force you to match up the "$n$"
in the formula in different ways with the number of terms in your series.
I chose the form above because things that you wrote imply that that
is the way it is written in your textbook.

One neat trick for dealing with the fact that someone might use a
letter such as $n$ to mean "number of terms" and someone else uses it
to mean "one less than the number of terms" is: use a different letter
for each meaning. In your case, it's just as valid to write about $Q_k$
for $k=0,1,2,ldots$ as $Q_n$ for $n=0,1,2,ldots.$
On the other hand, there's nothing to stop you from replacing $n$
with a different letter in the general closed series formula,
as long as you do the same thing on both sides of the equation.
That is, a perfectly good general closed form for a geometric series is
$$
a + ax + ax^2 + ax^3 + cdots + ax^m-1 = fraca(1-x^m)1-x.
$$
It's just like the previous formula, except with $m$ instead of $n.$

If your exercise literally used the letter $n$ in the expression $Q_n,$
you probably want to continue to use $n$ that way, so I'll assume we
want to use the general geometric series with $m$ instead.

To match up the formula correctly with the series in a particular problem,
it's usually sufficient to look at one or two simple examples.
In your problem, consider $Q_2$:
$$
Q_2=75+(75cdot 0.07)+(75cdot 0.07^2).
$$
The first term of the series is always $a$ in the general form,
so to apply the general formula to $Q_n$ we have to set $a=75.$
Each term should be $x$ times the previous one, but it's $0.07$ times,
so $x=0.07.$
Now observe that with $a=75$ and $x=0.07,$ the last term of $Q_n$
can be written $ax^n,$ but we want to use an equation whose last term
on the left-hand side is $ax^m-1.$ The "obvious" solution is to
set $m-1=n.$

But if $m-1=n$ then $m=n+1$; so what is $fraca(1-x^m)1-x$
in terms of the numbers and symbols you want to use?

Now to solve $P_n,$ let's forget all the particular definitions of
symbols such as $a,$ $x,$ and $m$ that are not in your original problem statement, and start over. As it turns out, some of the assignments of
values to these symbols that worked OK for $Q_n$ are not good for $P_n.$

For $P_n,$ there is also the following extra complication.
Consider $P_3,$ for example:
$P_3=(75cdot 0.07)+(75cdot 0.07^2)+(75cdot 0.07^3).$
Remember that in the general equation for the geometric series,
the first term is $a$; that means when you apply the equation to $P_n,$
you will need to set $a=75cdot 0.07.$
If you set $a=75,$ the first term of $P_n$ is $ax,$ not $a,$
and you can't simply apply the equation for the closed form.
So don't do that!

You still have $x=0.07,$ but $75cdot 0.07^2 = ax$ (not $ax^2$)
and $75cdot 0.07^3 = ax^2$ (not $ax^3$).
That's because
$$ax^3 = (75cdot 0.07)cdot x^3 = (75cdot 0.07)cdot 0.07^3 =
75cdot 0.07^4.$$
So we see that the subscript of $P$ doesn't match the exponent of $x$ in the
last term of the series form.
Instead, the last term of the series form of $P_n$ is $ax^n-1.$

I called this a "complication," but in one way it actually makes things
simpler for you, because it means that
$$
P_n = a + ax + ax^2 + ax^3 + cdots + ax^n-1,
$$
that is, when we match up the numbers in $P_n$ with the symbols in
the general series formulas, they come out just like
the left-hand side of the equation in your textbook.
So there's no need to change any symbols in that equation in order to
use it to get the closed form of $P_n.$

An even simpler way to deal with $P_n,$ however, is to look at
the series for $P_n$ and $Q_n,$ written out in a way that makes
it easy to compare them. For example,
beginalign
Q_3=&75+(75cdot 0.07)+(75cdot 0.07^2)+(75cdot 0.07^3),\
P_3=&phantom75+(75cdot 0.07)+(75cdot 0.07^2)+(75cdot 0.07^3).
endalign
It should be clear that in general, $P_n = Q_n - 75,$ so once you
have a closed form for $Q_n$ you can write a closed
form of $P_n$ in a few seconds.