
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 testtube, 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 testtube.

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)
Now we use your findings.
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.

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.

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.

Thanks a lot ticbol. Can I inquire where you are located?
Thanks also for spending so much time on this question.

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.

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.

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.

Sorry, MathGuru, but that is not chess.