Use induction to show the following for Fibonacci numbers:$ F_2 + F_4 + · · · + F_{2n} = F_{2n+1} − 1$ for every positive integer $n$

Question

Supposed to use induction, not really sure how to do this.

Answer 1

Hint: Induction works by first establishing the base case. First, show that this is true for $n=1$. Then you want to apply the inductive step, proving that if it holds for $n=k$, then it holds for $n=k+1$. So assume that:

$$f_2+f_4+...+f_{2k}=f_{2k+1}-1 \tag{1}$$

is true. Then you want to prove that:

$$f_2+f_4+...+f_{2k}+f_{2(k+1)}=f_{2(k+1)+1}-1 \tag{2}$$

using $(1)$. From $(1)$, we can add $f_{2(k+1)}=f_{2k+2}$ on both sides to get:

$$f_2+f_4+...+f_{2k}+f_{2(k+1)}=f_{2k+1}-1+f_{2k+2} \tag{3}$$

Now how can you transform the RHS of $(3)$ to get the RHS of $(2)$? I leave the rest to you (as a hint remember the definition of a Fibonacci number).

Answer 2

There are four steps to an induction problem.

Step $1$: Check the base case. This is the first number in the set you are working with (here positive integers), and, like in this case, it is almost always $n=1$. In other words, you should show that $F_{2\cdot 1}=F_{2\cdot1 +1}-1$, i.e. $F_2=F_3-1$.

Step $2$: Assume the condition holds for an arbitrary $k\in \Bbb Z^+$, that is, assume that $$\sum_{j=1}^kF_{2j}=F_{2k+1}-1$$ is true. (I have transposed your question into summation notation to make it easier to work with)

Step $3$: Manipulate the equation in step $2$ to show that the property will also hold for $n=k+1$. That is, show that

$$\sum_{j=1}^kF_{2j}=F_{2k+1}-1\implies\sum_{j=1}^{k+1}F_{2j}=F_{2(k+1)+1}-1$$

Step $4$: Write a conclusion explaining that, having shown the statement is true for $n=1$, and true for $n=k+1$ when $n=k$ has been assumed true, the statement must be true for all $n\in \Bbb Z^+$.

I've outlined what you need to achieve. How to achieve it is your task.