Mathematical Induction using a strong hypothesis

In summary, the conversation is about proving by induction that an=2^n for n>=0, given that a0=1, a1=2, and an=(a(n-1))^2/an-2 for n>=2. The conversation includes an attempt at solving the problem and a question about the inductive hypothesis. The conclusion is that the problem is easy and the solution provided appears to be correct.
  • #1
mamma_mia66
52
0

Homework Statement


If a0=1, and a1=2, and

an=(a(n-1))^2/an-2 for n>=2,
prove by induction that an=2^n for n>=0



Homework Equations





The Attempt at a Solution


(B) a0=1=2^0=1 yes is true
a1=2=2^1=2 yes is true

(I) ak=(2k-1)^2/2k-2=2k

Is it true that what I solved. It seems very easy.

I started first with a(k+1)=(a(k+1-1))^2/a(k+1-2)=
(2k)^2(k-1)=2(k+1)
Please someone help if I am doing something wrong. I will appreciate. Thank you.
 
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  • #2
It seems that you have the right idea, but I do not know what is different between (I) and your solution for ak+1. Where precisely is your inductive hypothesis?
 
  • #3
It IS very easy. So your induction step is: if a_{k-1}=2^(k-1) and a_{k-2}=2^(k-2) then a_k=2^(k) (by doing the algebra you doubtless did). It looks fine to me.
 
  • #4
That is where I am confused. I know that I need to fallow the form an=2^n

Then I don't need to do a(k+1)...

I did so many problems in my HW and now I don't get it what I am doing:smile:
 
  • #5
Thank you guys.
 

FAQ: Mathematical Induction using a strong hypothesis

What is mathematical induction using a strong hypothesis?

Mathematical induction using a strong hypothesis is a proof technique used to show that a statement is true for all natural numbers by first proving it for a specific case, and then showing that if it is true for any number, it must also be true for the next number.

What is the difference between strong and weak induction?

The difference between strong and weak induction lies in the initial assumption. In strong induction, we assume that the statement is true for all numbers up to and including the current number, whereas in weak induction we only assume that the statement is true for the previous number.

How do you use mathematical induction using a strong hypothesis?

To use mathematical induction using a strong hypothesis, you must first prove the statement for a base case. Then, assuming that the statement is true for all numbers up to and including a particular number, you must show that it is also true for the next number. This will prove that the statement is true for all natural numbers.

What are the advantages of using strong induction?

One advantage of using strong induction is that it allows us to prove statements that cannot be proven using weak induction. Additionally, strong induction can often lead to simpler and more elegant proofs compared to other proof techniques.

What are some common mistakes to avoid when using mathematical induction with a strong hypothesis?

One common mistake is to assume that the statement is true for all numbers without first proving it for a base case. Another mistake is to assume that the statement is true for the next number without showing that it is also true for all numbers up to and including that number. It is also important to use precise and clear language in the proof to avoid any confusion.

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