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i'm in a collage , studying sequence and series.

Prednisone is often prescribed for acute asthma attacks. For 5 mg tablets, typical instructions are: “Take 8 tablets the first day, 7 the second, and decrease by one tablet each day until all tablets are gone.” Prednisone decays exponentially in the body, and 24 hours after taking k mg, there are kx mg in the body.

(a) Write formulas involving x for the amount of prednisone in the body

the exponential decay is formed http://www.freemathhelp.com/forum/attachment.php?attachmentid=2602&stc=1

so i think ...

...........(i) 24 hours after taking the first dose (of 8 tablets), right before taking the second dose (of 7 tablets)

is the answer "8*5e^(-kx)" ? ....i don't know what k is.

...........(ii) Immediately after taking the second dose (of 7 tablets).

"8*5e^(-kx)+ 7*5e^(-kx)" ?

...........(iii) Immediately after taking the third dose (of 6 tablets).

"8*5e^(-kx)+ 7*5e^(-kx)+ 6*5e^(-kx)" ?

...........(iv) Immediately after taking the eighth dose (of 1 tablet).

"8*5e^(-kx)+ 7*5e^(-kx)+ 6*5e^(-kx) + ... + e^(-kx)" ?

...........(v) 24 hours after taking the eighth dose.

i don't know how to find it.

...........(vi) n days after taking the eighth dose.

i don't know how to find it.

(b) Find a closed form for the sum T = 8x^7+ 7x^6 + 6x^5 + ••• + 2x + 1, which is the number of prednisone tablets in the body immediately after taking the eighth dose.

Tn = nx^(n-1) ??

(c) If a patient takes all the prednisone tablets as prescribed, how many days after taking the eighth dose is there less than 3% of a prednisone tablet in the patient's body? The half-life of prednisone is about 24 hours.

i don't know how to find it.

(d) A patient is prescribed n tablets of prednisone the first day, n - 1 the second, and one tablet fewer each day until all tablets are gone. Write a formula that represents Tn, the number of prednisone tablets in the body immediately after taking all tablets. Find a closed form sum for Tn.

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closed as off-topic by Eevee Trainer, Lord Shark the Unknown, max_zorn, Shailesh, Riccardo.Alestra Jan 10 at 9:56

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    $\begingroup$ Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command ("Write..."), not a request for help, so please consider rewriting it. $\endgroup$ – Zev Chonoles Feb 9 '13 at 6:38
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Hint: The line $24$ hours after taking $k$ mg you have $kx$ mg gives you the answer to a). You have $40$ mg at the start, so you have $40x$ after $24$ hours. Then you take $35$ more, so you have what? Then a day wait multiplies by $x$ again, and so on.

for b: do you know how to sum the geometric series $1+x+x^2+x^3+\ldots x^8$? Do you see your series as the derivative of this.

For c: you are now given that $x=\frac 12$

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  • $\begingroup$ Thank you ... (((A))) i)40x ii)40x+35 iii)(40x+35)x + 30 = 40x^2 + 35x + 30 iv)40x^8 + 35x^7 + 30x^6 + ... $\endgroup$ – user61572 Feb 12 '13 at 8:39
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We do the "closed form" part, for the general case d). If $n$ pills are taken on the first day, $n-1$ on the second day, and so on, then immediately after the last pill there are $5(1+2x+3x^2+\cdots +nx^{n-1})$ milligrams in the body. Let's forget about the $5$, we can multiply by it at the end. So let
$$S(x)=1+2x+3x^2+\cdots +(n-1)x^{n-2}+ nx^{n-1}.$$

We will put $S(x)$ in a closed form probably close to what is expected. Note that $$xS(x)=x+2x^2+3x^3+\cdots+(n-1)x^{n-1}+ nx^n.$$ Subtract $xS(x)$ from $S(x)$. We get $$(1-x)S(x)=1+x+x^2+\cdots +x^{n-1}-nx^nx^8.\tag{$1$}$$ We may recognize $1+x+x^2+\cdots +x^{n-1}$ as the sum of a finite geometric series. This sum is $\frac{1-x^n}{1-x}$ (if $x\ne 1$). Thus from $(1)$ we see that $$(1-x)S(x)=\frac{1-x^n}{1-x}-nx^n.$$ Divide both sides by $1-x$. We get $$S(x)=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x}.$$ We can if we wish bring the above expression to the common denominator $(1-x)^2$.

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