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

Solution 1:

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?

Solution 2:

This is the core part of the inductive argument (I'll omit comments--try to piece together how each step works and comment if something is unclear): \begin{align} \sum_{i=1}^{k+1}f_{2i}&=\sum_{i=1}^kf_{2i}+f_{2k+2}\\[1em] &= (f_{2k+1}-1)+f_{2k+2}\\[1em] &= (f_{2k+2}+f_{2k+1})-1\\[1em] &= f_{2k+3}-1\\[1em] &= f_{2(k+1)+1}-1. \end{align}