1 Attachment(s)

Nth term in a geometric series

Attachment 27637So for this I have calculated what I believe to be the correct few terms of Q and P Q_{0}=75_{, }Q1=84, Q2=94.08. Always increasing by 12% of what the previous function was. I can't understand why my formula didn't work. I also tried using Q_{n}=Q_{0}*R^{n-1} but that provided an answer for n=0that was smaller than 75.

*Edit, I understand that I cant use n-1 because we're starting from n=0 and not n=1 so shouldn't it just be Q_{n}=Q_{0}*R^{n?}

Re: Nth term in a geometric series

If I understand the question correctly, then

$\displaystyle Q_0 = 75$

$\displaystyle Q_n= (.12)Q_{n-1} + 75$, in other words, what remains in the system on the previous day plus the new dose of 75mg per day.

EDIT:// this appears right in terms of a recursive equation, now all you would need to do is simplify to closed form.