7

Finding $\displaystyle \binom{2n}{1}^2-2\binom{2n}{2}^2+3\binom{2n}{3}^2-\cdots \cdots -2n\binom{2n}{2n}^2.$

What I've tried:

$$(1-x)^{2n}=\binom{2n}{0}-\binom{2n}{1}x+\binom{2n}{2}x^2+\cdots \cdots +\binom{2n}{2n}x^{2n}$$

$$-2n(1-x)^{2n-1}=-\binom{2n}{1}+2\binom{2n}{2}x-3\binom{2n}{3}x^2+\cdots +n\binom{2n}{2n}x^{2n-1}$$

Sum notation: $$\sum_{k=0}^{2n} (-1)^{k-1}k\binom{2n}{k}^2$$

Daniele Tampieri
  • 9,727
jacky
  • 5,194

3 Answers3

3

The solution uses the series representation of Legendre polynomials:

$P_n(x)=\frac{1}{2^n}\sum\limits_{k=0}^n \binom{n}{k}^2(x-1)^{n-k}(x+1)^k\tag1$

$\frac{x+1}{x-1}=-1$ at $x=0$ is valid. Extend the original sum (S) in the following way:

$S\frac{(x-1)^{2n}}{2^{2n}}=\frac{1}{2^{2n}}\sum\limits_{k=0}^{2n}\binom{2n}{k}^2 k(\frac{x+1}{x-1})^{k-1}(x-1)^{2n}\tag2$

We can realize that

$k(\frac{x+1}{x-1})^{k-1}=(-\frac{(x-1)^2}{2})\frac {d}{dx}(\frac{x+1}{x-1})^{k}\tag3$

Put it back to eqution (2) and replace the order of sum and derivation we get:

$S(x)\frac{(x-1)^{2n}}{2^{2n}}=-\frac{(x-1)^2}{2}\frac {d}{dx}\Big(\frac{1}{2^{2n}}\sum\limits_{k=0}^{2n}\binom{2n}{k}^2 (\frac{x+1}{x-1})^{k}(x-1)^{2n}\Big)\tag4$

Let's compare the summa part of (4) and $P_n(x)$ we get:

$S(x)\frac{(x-1)^{2n}}{2^{2n}}=-\frac{(x-1)^2}{2}\dfrac {dP_{2n}(x)}{dx}\tag5$

Finally

$S(x)=\frac{2^{2n-1}}{(x-1)^{2n-2}}\dfrac {dP_{2n}(x)}{dx}\tag6$

Applying the recursion relation of the Legendre polynomials we have at $x=0$:

$S(0)= 2n2^{2n-1}P_{2n-2}(0)\tag7$

JV.Stalker
  • 1,533
1

Starting from

$$\sum_{k=0}^{2n} (-1)^{k-1} k {2n\choose k}^2 = \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k} k {2n\choose k} \\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k} {2n-1\choose k-1} = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k} {2n-1\choose 2n-k} \\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k} [z^{2n-k}] (1+z)^{2n-1} \\ = 2n [z^{2n}] (1+z)^{2n-1} \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k} z^k \\ = 2n [z^{2n}] (1+z)^{2n-1} \left(1+\sum_{k=0}^{2n} (-1)^{k-1} {2n\choose k} z^k\right).$$

Now $2n [z^{2n}] (1+z)^{2n-1}$ is zero, so we may continue with

$$2n [z^{2n}] (1+z)^{2n-1} \sum_{k=0}^{2n} (-1)^{k-1} {2n\choose k} z^k \\ = - 2n [z^{2n}] (1+z)^{2n-1} (1-z)^{2n} = - 2n [z^{2n}] (1-z^2)^{2n-1} (1-z) \\ = - 2n [z^{2n}] (1-z^2)^{2n-1} = - 2n [z^{n}] (1-z)^{2n-1}.$$

This is

$$(-1)^{n+1} \times 2n \times {2n-1\choose n} = (-1)^{n+1} \times 2n \times {2n\choose n} \frac{n}{2n}$$

for an answer of

$$\bbox[5px,border:2px solid #00A000]{ (-1)^{n+1} \times n \times {2n\choose n}.}$$

Marko Riedel
  • 61,317
1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{k = 0}^{2n}\pars{-1}^{k - 1}\, k\, {2n \choose k}^{\!\! 2}} = \left.\partiald{}{x}\sum_{k = 0}^{2n}x^{k}{2n \choose k} {2n \choose 2n - k}\,\right\vert_{\ x\ =\ -1} \\[5mm] = &\ \left.\partiald{}{x}\sum_{k = 0}^{2n}x^{k}{2n \choose k} \bracks{z^{2n - k}}\pars{1 + z}^{2n}\,\right\vert_{\ x\ =\ -1} \\[5mm] = &\ \left.\bracks{z^{2n}}\pars{1 + z}^{2n}\, \partiald{}{x}\sum_{k = 0}^{2n}{2n \choose k}\pars{zx}^{k} \,\right\vert_{\ x\ =\ -1} \\[5mm] = &\ \left.\bracks{z^{2n}}\pars{1 + z}^{2n}\, \partiald{\pars{1 + zx}^{2n}}{x}\,\right\vert_{\ x\ =\ -1} \\[5mm] = &\ \bracks{z^{2n}}\pars{1 + z}^{2n}\, \bracks{2n\pars{1 - z}^{2n -1}\, z} \\[5mm] = &\ 2n\bracks{z^{2n - 1}}\bracks{% \pars{1 + z}^{2n - 1} + z\pars{1 + z}^{2n - 1}}\pars{1 - z}^{2n - 1} \\[5mm] = &\ 2n\braces{\underbrace{\bracks{z^{2n - 1}}\pars{1 - z^{2}}^{2n - 1}} _{\ds{\ =\ 0}}\ +\ \bracks{z^{2n - 2}}\pars{1 - z^{2}}^{2n - 1}} \\[5mm] = &\ 2n\bracks{{2n - 1 \choose n - 1}\pars{-1}^{n - 1}} = \bbx{\large\pars{-1}^{n - 1}\, n^{2}{2n \choose n}} \\ & \end{align}

Felix Marin
  • 89,464