Proving that $f_2+f_4+\cdots+f_{2n}=f_{2n+1}-1$ for Fibonacci numbers by induction

Given: $f_1 = f_2 = 1$ and for $n \in\mathbb{N}$, $f_{n+2} =f_{n+1} + f_n$.

Prove that $f_2 + f_4 + \dots + f_{2n} = f_{2n+1}- 1$.

Would you start with setting $f_2 + f_4 + \dots + f_{2n}= a_n$?

Then for the base case let $a_1=1$ LHS$=1$ and RHS$=2-1=1$ so base case holds.

Then the inductive hypothesis: Assume $f_2 + f_4 + \dots + f_{2n} = f_{2n+1}- 1$

$\textbf{NTS}$: $f_2 + f_4 + \dots + f_{2n} +f_{2n+2} = f_{2n+3}- 1$

Inductive step: By inductive hypothesis $f_2 + f_4 + \dots + f_{2n}=f_{2n+1}- 1$

So $f_{2n+1}- 1+f_{2n+1}$=$f_{2n+2}- 1$. As was to be shown.

Is this correct or did I need to show more algebra in my inductive step ?

• In your second line, you can just add the $-1$ over to the other side. The left-hand side collapses down to $f_{2n+1}$. – JasonM Jun 17 '16 at 4:31
• Possible duplicate of math.stackexchange.com/questions/787341/…. – lhf Jun 24 '16 at 0:57

Hint. The inductive step is rather $$f_2 + f_4 + \cdots + f_{2n}+\color{red}{f_{2n+2}}=\color{red}{f_{2n+3}}- 1,$$ then using the inductive hypothesis, we have to prove that $$f_{2n+1}-1+\color{red}{f_{2n+2}}=\color{red}{f_{2n+3}}- 1.$$ Can you take it from here?