What is the result of the sum of binomial coefficients with alternating signs?

[hxthanh]

Evaluate sum:

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

 

[Opalg]

Evaluate sum:

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

I believe that the answer must be \(\displaystyle S = \frac{(-1)^n(6n)!(2n)!}{(4n)!(3n)!n!}\), but I have NO idea how one might prove that.

What I did was to calculate $S_n$ when $n=1,\,2,\,3$, and then look those numbers up in the OEIS. This gave the above formula, and suggested that the next result, when $n=4$, would be $104006$. I then laboriously calculated $S_4$, and found that it is indeed $104006$. So my proposed formula has predictive value and therefore must be correct. (Wasntme) Now perhaps somebody can tell me why.

 

[hxthanh]

I believe that the answer must be \(\displaystyle S = \frac{(-1)^n(6n)!(2n)!}{(4n)!(3n)!n!}\), but I have NO idea how one might prove that.
...

Your result is absolute correct!(Clapping)
Hint: [sp] $(1-x^2)^m(1+x)^{2m}=(1-x)^m(1+x)^{3m}$[/sp]

Good luck! (Sun)

 

[hxthanh]

My solution

Spoiler

We start from the formula $(1-x^2)^m(1+x)^{2m}=(1-x)^m(1+x)^{3m}$
Now applying the binomial theorem
We have: $$ \sum_{k=0}^m (-1)^k{m\choose k}x^{2k} \sum_{j=0}^{2m} {2m\choose j}x^j = \sum_{k=0}^m (-1)^k{m\choose k}x^k \sum_{j=0}^{3m} {3m\choose j}x^j$$

\begin{equation} \label{eq1}\tag{1}\Leftrightarrow \sum_{k=0}^m \sum_{j=0}^{2m}(-1)^k {m\choose k} {2m\choose j}x^{2k+j} = \sum_{k=0}^m \sum_{j=0}^{3m} (-1)^k {m\choose k}{3m\choose j}x^{k+j} \end{equation}
compare coefficient of $x^{2m}$ in $\eqref{eq1} $ then we get
$\displaystyle \quad\sum_{2k+j=2m}(-1)^k {m\choose k} {2m\choose j}=\sum_{k+j=2m}(-1)^k {m\choose k} {3m\choose j}$
$\displaystyle \Leftrightarrow \sum_{k=0}^m(-1)^k {m\choose k} {2m\choose 2m-2k}=\sum_{k=0}^m (-1)^k {m\choose k} {3m\choose 2m-k}$
$\displaystyle \Leftrightarrow \sum_{k=0}^m(-1)^k {m\choose k} {2m\choose 2k}=\sum_{k=0}^m (-1)^k {m\choose k} {3m\choose m+k}$

with $m=2n$ then
\begin{equation} \label{eq2} \begin{aligned} S&=\sum_{k=0}^{2n}(-1)^k {2n\choose k} {4n\choose 2k}=\sum_{k=0}^{2n} (-1)^k {2n\choose k} {6n\choose 2n+k} \\ &=\sum_{k=0}^{2n} \dfrac{(-1)^k(2n)!(6n!)}{k!(2n-k)!(2n+k)!(4n-k)!} \\ &=\dfrac{(2n)!(6n)!}{(4n)!(4n)!} \sum_{k=0}^{2n} \dfrac{(-1)^k(4n)!(4n!)}{k!(4n-k)!(2n+k)!(2n-k)!} \\ &= \dfrac{(2n)!(6n)!}{(4n)!(4n)!} \sum_{k=0}^{2n} (-1)^k {4n\choose k} {4n\choose 2n-k} \\ &= \dfrac{(2n)!(6n)!}{(4n)!(4n)!} \sum_{k+j=2n} (-1)^k {4n\choose k} {4n\choose j}\tag{2} \end{aligned} \end{equation}
next to formula $(1-x^2)^{4r}=(1-x)^{4r}(1+x)^{4r}$
we get $ \displaystyle \sum_{k=0}^{4r}(-1)^k {4r\choose k}x^{2k}=\sum_{k=0}^{4r} \sum_{j=0}^{4r} (-1)^k {4r\choose k} {4r\choose j}x^{k+j}$
now, compare coefficient of $x^{2r}$ in that then we get
\begin{equation} \label{eq3}\tag{3} (-1)^r {4r\choose r}=\sum_{k+j=2r}(-1)^k {4r\choose k} {4r\choose j}\end{equation}
From $\eqref{eq2}$ and $\eqref{eq3}$, we have:
$$S=\dfrac{(2n)!(6n)!}{(4n)!(4n)!}\cdot(-1)^n{4n\choose n}=\dfrac{(-1)^n(2n)!(6n)!}{n!(3n)!(4n)!}$$

 

[CellCoree]

The sum given is a combination of binomial coefficients, which are commonly used in probability and statistics. The expression represents the sum of all possible combinations of choosing k objects from a set of 2n objects and then choosing 2k objects from a set of 4n objects. The alternating sign of (-1)^k indicates that the terms in the sum will alternate between positive and negative values.
To evaluate this sum, we can use the Binomial Theorem, which states that (a+b)^n can be expanded into a sum of binomial coefficients. In this case, a=1 and b=-1, so we can rewrite the expression as (1-1)^{6n}, which simplifies to 0. Therefore, the sum S is equal to 0.
This result may seem counterintuitive, as we would expect the sum to be a positive value. However, the alternating signs of the coefficients and the large number of terms in the sum cause the values to cancel each other out, resulting in a final value of 0. This can also be seen by considering the symmetry of the expression, where for every positive term there is a corresponding negative term.
In conclusion, the sum of binomial coefficients given by the expression is equal to 0. This result is important in understanding the properties of binomial coefficients and their applications in probability and statistics.