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The Fibonacci sequence $F_0, F_1, F_2, \ldots$ is defined recursively by $F_{0}:=0, F_{1}:=1 $ and $F_{n}:=F_{n-1}+F_{n-2}$.

Prove that $$\sum_{i=0}^{n} F_{i}=F_{n+2}-1 \qquad \text{for all } n \geq 0 .$$

I am stuck though on the way to prove this statement of fibonacci numbers by induction :

my steps:
definition:

The Hypothesis is: $\sum_{i=0}^{n} F_{i}=F_{n+2}-1$ for all $n > 1$

Base case: $n=2$
$\sum_{i=0}^{2} F_{i}=F_{0}+F_{1}+F_{2}=0+1+F_{1}+F_{0}=0+1+1+0=2$ which is equal to $F_{2+2}-1=F_{4}-1=F_{3}+F_{2}-1=F_{2}+F_{1}+F_{2}-1=1+1+1-1=2$ OK!

inductive step:
to prove: $\sum_{i=0}^{n+1} F_{i}=F_{n+3}-1$ for all $n > 1$
$\sum_{i=0}^{n+1} F_{i}=\sum_{i=0}^{n} F_{i}+F_{n+1}=F_{n+2}-1+F_{n+1}=...help...=F_{n+3}-1$

i need help to $..help..$ please! thanks a lot

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Use $F_{n+1}+F_{n+2}=F_{n+3}$, to get:

$$\sum_{i=0}^{n+1} F_{i}=\sum_{i=0}^{n} F_{i}+F_{n+1}=F_{n+2}-1+F_{n+1}=F_{n+1}+F_{n+2}-1=F_{n+3}-1$$

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  • $\begingroup$ why $(n+2)+(n+1) = n+3 $ in fibobacci ? $\endgroup$
    – doniyor
    Nov 24 '12 at 8:52
  • $\begingroup$ So what? Stil $F_{n+3}=F_{n+2}+F_{n+1}$ holds. $\endgroup$
    – Amr
    Nov 24 '12 at 8:54
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    $\begingroup$ $F_{n + k + 2} = F_{n + k + 1 } + F_{n + k}$ holds true always in the Fibonacci Sequence, as long as $n$ and $k$ are whole numbers. You could even remove the $k$ and get the correct definition. The $k$ was pushed in for better understanding. You could put infinite $k$'s in there—i.e., $k_1 , k_2\ldots$. $\endgroup$
    – P.K.
    Nov 24 '12 at 9:55

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