A strain of bacteria doubles every $14$ h. If there are $100$ bacteria cells to start with in a colony, how many will there be in $7$ days?

This is a sequence question.

My answer: We start with $100$ cells, so $a = 100$. We double, so $\text{common ratio} = 2$. The formula will be

$$100\cdot 2^{n-1}=100\cdot 2^{14-1}=100\cdot 2^{13}$$

My answer sheet says I'm incorrect, as the answer is $100\cdot 2^{14}$, not $100\cdot 2^{13}$, why is this so? How do I know if I must use the exponent $n-1$, or $n$? I'm confused. Some examples end up being to the $n$-th power, and some end up being to the $(n-1)$-th power.

I'm having a basically identical problem to this person: How To Know in a Application Sequence/Series Problem Which Variable is $a_{0}$ or $a_{1}$?

Thank you!


Both your answer sheet and you are wrong. $7$ days is $168$ hours, which is $12$ periods of $14$ hours, so the number doubles $12$ times. The number at the end is then $100 \cdot 2^{12}$

To tell whether it is $n$ or $n-1$, you need to think about how the problem is stated. The ratio is applied in the gaps, so think about how many gaps there are. If the problem says "on the first day ...", you start counting with $1$. If you count up to $n$, there are $n-1$ gaps, so the power will be $n-1$. In this case, we are told that at the starting moment there are $100 \cdot 2^0$ bacteria and then we go through $12$ intervals.

  • $\begingroup$ What about say for this question?: Boxes are stacked in a store display in the shape of a triangle. The number of boxes in the rows form an arithmetic sequence. There are 41 boxes in the 3rd row from the bottom. There are 23 boxes in the 12th row from the bottom. What is the formula for the nth term of the sequences? The ratio isn't really applied in the gaps. Its just added whenever we reach a new row. $\endgroup$ – user164403 Jun 16 '15 at 23:41
  • $\begingroup$ In this case it is not a ratio, it is an addition. The gaps are the spaces between the rows where the addition is applied. It is the same idea. $\endgroup$ – Ross Millikan Jun 17 '15 at 0:19
  • $\begingroup$ Could you give an example of when there is just d * n, as opposed to * (n-1)? Perhaps if we were to start from the "0th" day? Wouldn't in this case the bottom be the "0", as opposed to the "1"? So it'd be n instead of n-1? $\endgroup$ – user164403 Jun 17 '15 at 1:10
  • $\begingroup$ That is correct. If you call the first term $a_0$, then the general term is $a_n=a_0+nd$ $\endgroup$ – Ross Millikan Jun 17 '15 at 2:05
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    $\begingroup$ I would say so, but I think it is better to think about each problem than to try and come up with rules about the types of problem. $\endgroup$ – Ross Millikan Jun 17 '15 at 2:40

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