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I'm looking for some hints on proving the following (either directly or by induction):

$$ \sum_{k={0}}^{l/2} \frac{(-1)^k(2l-2k)!}{k!(l-k)!(l-2k)!} =2^l $$

I do know it is actually true from various sources, none of which have bothered to show me how it is done, so I wanted to try to prove it on my own. However, it seems to be out of the range of my current mathematical ability.

I tried using some blasphemous form of combinatorics and subsequently Pascal's Identity to simplify this and it did not end well:

$$ = \sum_{k={0}}^{l/2} (-1)^k \frac{(2l)!}{(l!)^2} {2l-2k \choose l} {2l-k \choose l} {2l \choose k} = \ldots = ? $$

I'm also not even sure how to prove this by induction by assuming it is true, trying the base case:

$$ l=0 \rightarrow \sum_{k={0}}^{l/2} \frac{(-1)^k(2l-2k)!}{k!(l-k)!(l-2k)!} = 2^l=2^0 $$

and then incrementing the assumed true equation from $l$ to $l+1$ $$ l=l+1 \rightarrow \sum_{k={0}}^{(l+1)/2} \frac{(-1)^k(2(l+1)-2k)!}{k!((l+1)-k)!((l+1)-2k)!} = 2^{l+1}\sum_{k={0}}^{l/2} \frac{2l+1-2k}{l+1-2k} \neq 2^{l+1} $$

I thought about using the Gamma function as well, but did not try that yet. Did I make a wrong assumption somewhere? I'm pretty sure I am missing something simple, or the solution is way out of left-field. Thoughts?

rlarson
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1 Answers1

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Notice your expression can be rewritten as

$$\sum_{k=0}^{\lfloor l/2\rfloor}{(-1)^k(2l-2k)!\over k!(l-k)!(l-2k)!} =\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k{n\choose k}{2n-2k\choose n}.$$

Consider the Legendre polynomials, which we choose to define by Rodrigues' formula as $$P_n(x)={1\over2^nn!}{d^n\over dx^n}(x^2-1)^n;$$ we may compute $$\begin{align*} P_n(x)&={1\over2^nn!}{d^n\over dx^n}\sum_{k=0}^n{n\choose k}(-1)^kx^{2n-2k}\\ &={1\over2^nn!}\sum_{k=0}^n{n\choose k}(-1)^k{d^n\over dx^n}x^{2n-2k}\\ &={1\over2^nn!}\sum_{k=0}^{\lfloor n/2\rfloor}{n\choose k}(-1)^k{(2n-2k)!\over(n-2k)!}x^{n-2k}\\ &={1\over2^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k{n\choose k}{2n-2k\choose n}x^{n-2k}. \end{align*}$$ Our goal now is thus to prove that $P_n(1)=1$. By the general Leibniz rule $${d^n\over dx^n}(fg)=\sum_{k=0}^n{n\choose k}{d^{n-k}f\over dx^{n-k}}{d^k g\over dx^k}$$ applied with $f=(x-1)^n$ and $g=(x+1)^n$ (so that $fg=(x^2-1)^n$), we have $$\begin{align*} P_n(x)&={1\over2^nn!}{d^n\over dx^n}(x^2-1)^n\\ &={1\over2^nn!}\sum_{k=0}^n{n\choose k}{n!\over k!}(x-1)^k{n!\over(n-k)!}(x+1)^{n-k}\\ &={1\over2^n}\sum_{k=0}^n{n\choose k}^2(x-1)^k(x+1)^{n-k}. \end{align*}$$ When evaluated at $x=1$, only the $k=0$ term survives, and it follows that $P_n(1)=1$ as needed.

ho boon suan
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