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I believe the majority of my proof is correct I'm just not certain about the base case if any one can explain how to do that base case or fix any error I made I would greatly appreciate it.

Recall the fibonacci sequence is defined by $f_1=f_2=1$ and for $n\in \mathbb{N}$, $f_n+f_{n+1}=f_{n+2}$ Prove that for every natural number $n$ that:$$f_1+f_3+\dots+f_{2n-1}=f_{2n}$$

By Induction

Let $a_n=f_1+f_3+\dots+f_{2n-1}$

Base case:Let $a_1=1$ Thus LHS$=1$ and RHS$=1$. Therefore the base case holds.

Induction Hypothesis: Assume $f_1+f_3+\dots+f_{2n-1}=f_{2n}$

NTS:$f_1+f_3+\dots+f_{2n-1}+f_{2n+1}=f_{2n+2}$

Inductive Step: By Induction Hypothesis the above simplifies to $f_{2n}+f_{2n+1}=f_{2n+2}$.

As was to be shown.

user348547
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1 Answers1

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Problems in Your Post

  • You wrote,

    By Induction,

    Let $a_n=f_1+f_3+⋯+f_{2n−1}$

    It is not clear from what you wrote what actually you are trying to prove. For example, you haven't defined what your $a_n$ is. You also haven't properly written what your actual statement is on which you are trying to apply induction. More precisely, to apply induction we need to have some statement $P(n)$ but in your case you haven't explicitly stated what your $P(n)$ is. Unless you state this, you can't apply the principle of mathematical induction.

  • Next you wrote,

    Base case: Let $a_1=1$. Thus LHS=$1$ and RHS=$1$. Therefore the base case holds.

    There are at least two problems in this statement. Remember that in the Base Case you are to prove that when in your statement $n=1$ (or, in general a suitable initial value) then $P(1)$ is true. But here how are you proving that when $a_1=1$ LHS=RHS? Furthermore, how do you even know that $a_1=1$?

What Your Proof Could Be

Define, $$P(n):=f_1+f_3+\ldots+f_{2n-1}=f_{2n}$$

Base Case: When $n=1$, in LHS we only have $f_1=1$ and in RHS we have $f_2=1$. So in this case LHS=RHS.

Induction Hypothesis: When $n=k$, $P(n)$ is true, i.e., $P(k)$ is true. In other words, $$f_1+f_3+\ldots+f_{2k-1}=f_{2k}$$ (What would happen if we would use the symbol $n$ instead of $k$?)

The remaining part of your proof seems fine to me although it could be made more explicit.