1. ## Bacteria by Induction

I am trying to prove this by induction, but have so far failed.

I note that:

b1 = 1, r1 = 7
b2 = 7, r2 = 13
b3 = 13, r3 = 55

But from there i am stuck. As always thanks to all.

A new species of bacteria called blue and red behave in the following way: after one minute every blue bacterium turns into 6 red ones, and every red bacterium turns into 1 blue and 1 red bacterium. During an experiment you put 1 blue and 1 red bacterium into a test-tube, so that after one minute there were 1 blue bacterium and 7 red ones; and after two minutes there were 7 blue ones and 13 red ones. Prove that after n minutes there will be:

3/5 * 3^n + 2/5 * (-2)^n
Blue bacteria in the test-tube.

2. Here is one way.

Prove by induction:
bn = (3/5)[3^n] +(2/5)[(-2)^n]

Okay.
Let us see first if b1 is true.
Since you found b1 = 1, we see if that is true.
When n = 1,
b1 = (3/5)[3^1] +(2/5)[(-2)^1]
b1 = (3/5)[3] +(2/5)[-2]
b1 = 9/5 -4/5
b1 = 5/5 = 1 ----it is true.

Then let us assume when n = k,
bk = (3/5)[3^k] +(2/5)[(-2)^k] is true.

Then if we can show that when n = k+1,
b_(k+1) = (3/5)[3^(k+1)] +(2/5)[(-2)^(k+1)] is true also,

then we can say that
bn = (3/5)[3^n] +(2/5)[(-2)^n] is really true.
It is proven by mathematical induction.

So,
b_(k+1) = (3/5)[3^(k+1)] +(2/5)[(-2)^(k+1)]
b_(k+1) = (3/5)[3^k *3^1] +(2/5)[(-2)^k *(-2)^1]
b_(k+1) = (3/5)[3^k *3] +(2/5)[(-2)^k *(-2)]
b_(k+1) = (9/5)[3^k] -(4/5)[(-2)^k] ----(i)

You found
b2 = 7
b_(2+1) = b3 = 13

Plugging those into the Eq.(i), when k=2,
b_(k+1) = (9/5)[3^k] -(4/5)[(-2)^k] ----(i)
b_(2+1) = (9/5)[3^2] -(4/5)[(-2)^2]
b_(2+1) = (9/5)[9] -(4/5)[4]
b_(2+1) = 81/5 -16/5
b_(2+1) = 65/5 = 13
Or,
b3 = 13 ---which is true.

Therefore, bn = (3/5)[3^n] +(2/5)[(-2)^n] is proven.

3. Very much appreciated ticbol!

However there is one thing which i want to clear up, in the end when you sub in b2 = 7, can you do that to prove it for all cases, i was wondering if that only proved it for the 2nd case, and not the kth + 1 case.

Any clarification would be great.

4. I tried not using your findings but I could not find a way.
After
b_(k+1) = (9/5)[3^k] -(4/5)[(-2)^k] ----(i)
I was stuck.
If bn were only a summation or sum, then maybe I could have found a way.
But since bn is defined differently, I was forced to use two of your findings.

By the way, I found also your findings by brute force, by diagrams.

Tomorrow, when I have time, I will try again.

----------
Umm, I think I missed your point.
So, you were asking if my using k=2 will cover all k+1?

Sure it will.
Remember, if k=2, then k+1 = 2+1 = 3
So if b3 is true by using k=2, then k and k+1 are satisfied.
We only need to satisfy one k+1 for the induction.

If you like, test this case.
k=1, and so k+1 = 2.
You will get b2 = 7 by using k=1.

Or,
k=3, anmd so k+1 = 4.
You will find b4 = 55 by using k=3.

5. Thanks a lot ticbol. Can I inquire where you are located?

Thanks also for spending so much time on this question.

6. ## Thank you ticbol

Ticbol thank you for providing such detailed answers you are a real asset to all of the members!

Please let me invite you both to our introductions thread if you care to share with us any information about yourselves.

7. Umm.

Well, I am a Filipino civil engineer working in constructions here in Guam.

I love Math. Like chess, it is one way for me to unwind.

8. ## The big knight switch

Like chess huh?

Have you seen The Big Knight Switch ?

I have spent about 20 minutes on it and I haven't solved it yet.

9. Sorry, MathGuru, but that is not chess.