[Math] Prove the sum of the even Fibonacci numbers

fibonacci-numberssummation

Let $f_n$ denote the $nth$ Fibonacci number. Prove that

$f_2\:+\:f_4\:+…+f_{2n}=f_{2n+1}-1$

I am having trouble proving this. I thought to use induction as well as Binet's formula where,

$f_n=\frac{\tau^2-\sigma^2}{\sqrt5}$ where $\tau=\frac{1+\sqrt5}{2}$ and $\sigma=\frac{-1}{\tau}$.

Can someone give me a hand?

Thanks!

Best Answer

Using the recurrence relation for $f_n$ we find

\begin{align}f_2 + f_4 + \cdots + f_{2n} = (f_3 - f_1) + (f_5 - f_3) + \cdots + (f_{2n+1} - f_{2n-1}), \end{align}

which telescopes to $f_{2n+1} - f_1 = f_{2n+1} - 1$.

Related Solutions

Nth Fibonacci number divided by 2 to the n

The generating function of $F_{2n}$ (as defined here) is $$\frac{x-x^2}{x^2-3x+1}=\sum_{n=1}^\infty F_{2n}x^n$$ Substitute $x=\frac14$ to get the sum as $$\sum_{n=1}^\infty\frac{F_{2n}}{2^{2n}}=\frac{1/4-1/16}{1/16-3/4+1}=\frac35$$

Zeckendorf with Negative Fibonacci Numbers

The negative fibbonacci numbers run as $F(-n)=F(n)$ for odd $n$ and $F(-n)=-F(n)$ for even $n$.

So we get for fibonacci numbers

 -21  13  -8   5  -3   2  -1   1
                          -1
                  -3           1
                  -3
                  -3      -1
          -8           2       1
          -8           2
          -8       t0 12
  -21          5       2       1
  -21     -8       -3     -1

So yes, all of the natural numbers can be expressed as a sum of non-consecutive fibonacci numbers of negative index.

It ought be a straight-forward conversion from the positive to negative schema.

  eg  -21            5     2      1
       21           -5    -2     -1
           13   8   -5    -2    -1
           13           3 -2     -1
           13

It's possible to go both ways, and there is a carry rule to prevent adjacent columns being filled with more than one counter.

Best Answer

Related Solutions

Nth Fibonacci number divided by 2 to the n

Zeckendorf with Negative Fibonacci Numbers

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