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The Fibonacci polynomials Fn(x) are defined by
Fn+1(x)=xFn(x)+Fn−1(x)n≥1F0(x)=0,F1(x)=1
They have generating function
G(x,t)=n≥0∑Fn(x)tn=1−xt−t2t
Expanding this generating function we obtain, in particular, that
F2n+1(x)=j=0∑n(n−jn+j)x2j
and hence the question asks us to evaluate F2n+1(2i). Now
G(2i,t)=1−2it−t2t=(1−it)2t=m≥0∑(m+1)imtm+1
so that
F2n+1(2i)=(2n+1)i2n=(−1)n(2n+1)
Sure, and we can do the third one by replacing k+1(−1)k by xk, identifying the new series as a 3F2 hypergeometric at x, and integrating that result to obtain the desired series as the value of a 4F3 terminating hypergeometric at unity, and then throwing a bundle of hypergeometric identities at the problem. It would be nice to find a combinatoric proof instead.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
The Fibonacci polynomials Fn(x) are defined by Fn+1(x)=xFn(x)+Fn−1(x)n≥1F0(x)=0,F1(x)=1 They have generating function G(x,t)=n≥0∑Fn(x)tn=1−xt−t2t Expanding this generating function we obtain, in particular, that F2n+1(x)=j=0∑n(n−jn+j)x2j and hence the question asks us to evaluate F2n+1(2i). Now G(2i,t)=1−2it−t2t=(1−it)2t=m≥0∑(m+1)imtm+1 so that F2n+1(2i)=(2n+1)i2n=(−1)n(2n+1)
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Nice! In fact this can be very easily converted into Gauss hypergeometric.
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Sure, and we can do the third one by replacing k+1(−1)k by xk, identifying the new series as a 3F2 hypergeometric at x, and integrating that result to obtain the desired series as the value of a 4F3 terminating hypergeometric at unity, and then throwing a bundle of hypergeometric identities at the problem. It would be nice to find a combinatoric proof instead.
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Yes. I agree, that's why I posted them here. BTW, there is a better way to get the third one.
Denote the LHS as An. Using the identity (rn)+(r−1n)=(rn+1) and re-indexing wherever required, it is easy to show that,
An+1+An−1+2An=0
Using the initial values A0=1, A1=−3, it follows that An=(−1)n(2n+1).
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Nice. Probably you would like this.