A bacteria colony begins with $6$ individuals and doubles in size every hour. Write down a recurrence for a population at the beginning of hour n, and solve it. How many hours elapse until the population exceeds one million?

This is what I wrote: $a_n=2a_{n-1}$ for $\forall n\ge 1$ where our $a_0=6$ (sorry I don't know how to write it properly)

However the solution continues with : We know $a_0 = 1$, so: $a_n = 2a_{n−1} = 2^2a_{n−2} = 2^3a_{n−3} = \cdots = 2^na_0 = 6\cdot 2^n$. This is the part which I do not understand. Can someone explain it to me?

Thank you!

  • $\begingroup$ I've edited it. Is that what you mean? $\endgroup$ – tarit goswami Apr 10 at 10:00
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    $\begingroup$ I think the solution was supposed to continue with "We know $a_0 = 6$". $\endgroup$ – Arthur Apr 10 at 10:07
  • $\begingroup$ tarit actually the solution was an = 2an−1 = 22an−1 = 23an−2 = ... = 2na0 = 6·2n. which may be wrong $\endgroup$ – Flea Apr 10 at 10:20

We know $a_0=1$

Here $a_0=6$. $a_n=2a_{n-1}$ is called recurrence relation, where $n$th term is related to $(n-1)$ th term by the given relation. Here $n$ is a variable which can take values $\ge 1$. Hence, we can also write $a_{n-1}=2a_{n-2}$ and $a_{n-2}=2a_{n-3}$ and so on upto $a_1=2a_0$(minimum value of $n$ can be $1$).

So, putting the values back, we can say, $$a_n=2a_{n-1}=2\cdot (2a_{n-2})=2^2a_{n-2}\\=2^2\cdot (2a_{n-3})=2^3a_{n-3}$$ and in general, $a_n=2^ka_{n-k}$ where $1\le k\le n$. So, finally you will have $a_n= 2^na_0=6\cdot 2^n$, as $a_0=6$. Have you understood now?

  • $\begingroup$ So the solution given by my teacher is wrong right? $\endgroup$ – Flea Apr 10 at 10:19
  • $\begingroup$ @Flea What was your teacher's solution? $\endgroup$ – tarit goswami Apr 10 at 10:22
  • $\begingroup$ How did you come up with an=2^na0=6*2^n? $\endgroup$ – Flea Apr 10 at 10:22
  • $\begingroup$ an = 2an−1, n ≥ 1 We know a0 = 1, so: an = 2an−1 = 2^2an−1 = 2^3an−2 = ... = 2na0 = 6·2n. Therefore: an = 6·2n ∀n ≥ 0. $\endgroup$ – Flea Apr 10 at 10:22
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    $\begingroup$ @Flea Welcome :) If it explained your query, you can accept this question by clicking the tick mark left of the answer $\endgroup$ – tarit goswami Apr 10 at 10:43

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