# Sum of square of binomial coeffcient with positive and negative terms

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

• I suspect at typo: $n$ in first $3$ terms of first line must be $2n$, right? – drhab Feb 15 at 9:40
• Try writing the sum in sigma notation first – unseen_rider Feb 15 at 9:58
• This looks right, now just substitute $x = 0$ (and fix the typo whereas $+n\binom{2n}{2n}x^{2n-1}$ should be $+2n\binom{2n}{2n}x^{2n-1}$). – Andreas Caranti Feb 15 at 13:17
• @Andreas Caranti, thanks, I am in a hurry, regards. – JV.Stalker Feb 15 at 13:27

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

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

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

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 $$x=0$$:

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