Fibonacci sequence in a double sum

Asked May 15 '19 at 10:43

Active May 15 '19 at 10:43

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$$\sum_{j=0}^{n}\sum_{k=0}^{n}{n-j \choose j}{n-k \choose k}=x$$

Setting $n:=1,2,3,4,5$ we find that $x:=1,1,4,9,25$

It looks like $\sqrt{x}:=1,1,2,3,5$ is the sequence of Fibonacci number.

I can't show that

$$\sum_{j=0}^{n}\sum_{k=0}^{n}{n-j \choose j}{n-k \choose k}=F_n^2$$

Where $F_n$ is the Fibonacci number

Am I correct that this double sum produces the Fibonacci sequence?

asked May 15 '19 at 10:43

1

That follows directly from https://math.stackexchange.com/questions/81805/how-to-show-that-this-binomial-sum-satisfies-the-fibonacci-relation, by squaring the identity $\sum_{k\ge0} \binom{n-k}k=F_{n+1}$. – Martin R May 15 '19 at 10:55
4

Possible duplicate of How to show that this binomial sum satisfies the Fibonacci relation? – Martin R May 15 '19 at 10:56

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