Are Binomial Coefficients and Fibonacci Numbers Related?

[ramsey2879]

I posted a bit about the characteristic value of a Fibonacci series without proof but nowI wonder about whether it could be valid. A remarkable result of it applies to the binominal coefficients. The sum of following Binominal coefficients times corresponing Fibonacci numbers are in them selves terms from the Fibonnacci series. In other words, each of the following sums is a Fibonacci number. "(n,i)" is used her to represent the the binominal coefficients

A. \sum_{i=0}^{n}F_{i-1}(n,i)

B. \sum_{i=0}^{n}F_{i-2}(n,i)

For intance;
F_{-1}*1 + F_{0}* 3 + F_{1}*3 + F_{2}*1 = F_{5}

F_{-2}*1 + F_{-1}*3 + F_{0}*3 + F_{1}*1 = F_{4}

F_{-1}*1 + F_{0}*4 + F{1}*6 + F_{2}*4 + F_{3}*1 = F_{7}

I could go on but that in it self does not prove the result.

 

[MathematicalPhysicist]

The only thing I can advise you is to check in Combinatorial Identities of John Riordan and another one by Henri Gould.

 

[ramsey2879]

The only thing I can advise you is to check in Combinatorial Identities of John Riordan and another one by Henri Gould.

Thanks but I thought of a proof by Induction
To Prove

F_{x+2n) = \sum_{i=0}^{n}F_{x+i}nCi

It is easily verified for n=1

If true for any positive n then it is true for n+1

F_{(x+2*(n+1))} = F_{(x+2n)} + F_{(x+2n+1)}

=\sum_{i=0}^{n}F_{x+i}nCi + F_{x+1+i}nCi

Combining like F terms and using (n+1)Ci = nC(i-1)+nCi gives the required result

 

[John Creighto]

I posted a bit about the characteristic value of a Fibonacci series without proof but nowI wonder about whether it could be valid. A remarkable result of it applies to the binominal coefficients. The sum of following Binominal coefficients times corresponing Fibonacci numbers are in them selves terms from the Fibonnacci series. In other words, each of the following sums is a Fibonacci number. "(n,i)" is used her to represent the the binominal coefficients

A. \sum_{i=0}^{n}F_{i-1}(n,i)

B. \sum_{i=0}^{n}F_{i-2}(n,i)

For intance;
F_{-1}*1 + F_{0}* 3 + F_{1}*3 + F_{2}*1 = F_{5}

F_{-2}*1 + F_{-1}*3 + F_{0}*3 + F_{1}*1 = F_{4}

F_{-1}*1 + F_{0}*4 + F{1}*6 + F_{2}*4 + F_{3}*1 = F_{7}

I could go on but that in it self does not prove the result.

How about using Pascel's triangle.

 

[John Creighto]

Anyway, I don't think it should be hard to prove using induction given that Fibonacci series is defined recursively and polynomial multiplication can be done via convolution.