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Binomial Coefficient Challenge 2!

$\displaystyle \sum_{k=0}^n{\binom{n+k}{2k}{(-4)}^k} = {(-1)}^{n}(2n+1)$

Note by Kartik Sharma
9 months ago

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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 $$A_{n} = (-1)^n (2n+1)$$.

- 8 months, 3 weeks ago

Nice. Probably you would like this.

- 8 months, 3 weeks 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 $F_{2n+1}(2i) \; = \; (2n+1)i^{2n} \; = \; (-1)^n (2n+1)$

- 8 months, 3 weeks ago

Nice! In fact this can be very easily converted into Gauss hypergeometric.

- 8 months, 3 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.

- 8 months, 3 weeks ago

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

- 8 months, 3 weeks ago