Denote the LHS as \(A_{n}\). Using the identity \(\dbinom{n}{r} + \dbinom{n}{r-1} = \dbinom{n+1}{r} \) and re-indexing wherever required, it is easy to show that,

\[A_{n+1} + A_{n-1} + 2A_{n} = 0\]

Using the initial values \(A_{0} = 1\), \(A_{1} = -3\), it follows that \(A_{n} = (-1)^n (2n+1)\).
–
Ishan Singh
·
2 months, 1 week ago

The Fibonacci polynomials \(F_n(x)\) are defined by
\[ F_{n+1}(x) \; = \; xF_n(x) + F_{n-1}(x) \hspace{1cm} n \ge 1 \hspace{3cm} F_0(x) \; = \;0\,,\,F_1(x) \; = \; 1 \]
They have generating function
\[ G(x,t) \; = \; \sum_{n \ge 0} F_n(x)t^n \; = \; \frac{t}{1 - xt - t^2} \]
Expanding this generating function we obtain, in particular, that
\[ F_{2n+1}(x) \; = \; \sum_{j=0}^n {n+j \choose n-j} x^{2j} \]
and hence the question asks us to evaluate \(F_{2n+1}(2i)\). Now
\[ G(2i,t) \; = \; \frac{t}{1 - 2it - t^2} \; = \; \frac{t}{(1 - it)^2} \; = \; \sum_{m \ge 0} (m+1)i^m t^{m+1} \]
so that
\[ F_{2n+1}(2i) \; = \; (2n+1)i^{2n} \; = \; (-1)^n (2n+1) \]
–
Mark Hennings
·
2 months, 2 weeks ago

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@Mark Hennings
–
Nice! In fact this can be very easily converted into Gauss hypergeometric.
–
Kartik Sharma
·
2 months, 2 weeks ago

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@Kartik Sharma
–
Sure, and we can do the third one by replacing \(\frac{(-1)^k}{k+1}\) by \(x^k\), identifying the new series as a \({}_3F_2\) hypergeometric at \(x\), and integrating that result to obtain the desired series as the value of a \({}_4F_3\) 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.
–
Mark Hennings
·
2 months, 2 weeks ago

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@Mark Hennings
–
Yes. I agree, that's why I posted them here. BTW, there is a better way to get the third one.
–
Kartik Sharma
·
2 months, 2 weeks ago

## Comments

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TopNewestDenote the LHS as \(A_{n}\). Using the identity \(\dbinom{n}{r} + \dbinom{n}{r-1} = \dbinom{n+1}{r} \) and re-indexing wherever required, it is easy to show that,

\[A_{n+1} + A_{n-1} + 2A_{n} = 0\]

Using the initial values \(A_{0} = 1\), \(A_{1} = -3\), it follows that \(A_{n} = (-1)^n (2n+1)\). – Ishan Singh · 2 months, 1 week ago

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this. – Kartik Sharma · 2 months, 1 week ago

Nice. Probably you would likeLog in to reply

The Fibonacci polynomials \(F_n(x)\) are defined by \[ F_{n+1}(x) \; = \; xF_n(x) + F_{n-1}(x) \hspace{1cm} n \ge 1 \hspace{3cm} F_0(x) \; = \;0\,,\,F_1(x) \; = \; 1 \] They have generating function \[ G(x,t) \; = \; \sum_{n \ge 0} F_n(x)t^n \; = \; \frac{t}{1 - xt - t^2} \] Expanding this generating function we obtain, in particular, that \[ F_{2n+1}(x) \; = \; \sum_{j=0}^n {n+j \choose n-j} x^{2j} \] and hence the question asks us to evaluate \(F_{2n+1}(2i)\). Now \[ G(2i,t) \; = \; \frac{t}{1 - 2it - t^2} \; = \; \frac{t}{(1 - it)^2} \; = \; \sum_{m \ge 0} (m+1)i^m t^{m+1} \] so that \[ F_{2n+1}(2i) \; = \; (2n+1)i^{2n} \; = \; (-1)^n (2n+1) \] – Mark Hennings · 2 months, 2 weeks ago

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– Kartik Sharma · 2 months, 2 weeks ago

Nice! In fact this can be very easily converted into Gauss hypergeometric.Log in to reply

– Mark Hennings · 2 months, 2 weeks ago

Sure, and we can do the third one by replacing \(\frac{(-1)^k}{k+1}\) by \(x^k\), identifying the new series as a \({}_3F_2\) hypergeometric at \(x\), and integrating that result to obtain the desired series as the value of a \({}_4F_3\) 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.Log in to reply

– Kartik Sharma · 2 months, 2 weeks ago

Yes. I agree, that's why I posted them here. BTW, there is a better way to get the third one.Log in to reply