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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 ?

Community
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user348165
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  • 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 04:31
  • Possible duplicate of http://math.stackexchange.com/questions/787341/summation-of-fibonacci-numbers-f-n-with-n-odd-vs-even. – lhf Jun 24 '16 at 00:57

2 Answers2

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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?

Olivier Oloa
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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}

Daniel W. Farlow
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