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Find $$\binom{n}{0} \binom{2n}{n}-\binom{n}{1} \binom{2n-2}{n}+\binom{n}{2} \binom{2n-4}{n}+\cdots$$

I have taken $r$th term and modified as follows:

$$T_r =(-1)^r \binom{n}{r} \binom{2n-2r}{n}=(-1)^r \frac{n!}{(n-r)!r!} \times \frac{ (2n-2r)!}{n! (n-2r)!}=(-1)^r \frac{(2n-2r)!}{(n-r)! r!(n-2r)!}$$

Can we continue from here?

Ekaveera Gouribhatla
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2 Answers2

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Here is an answer based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g. \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n(-1)^k}&\color{blue}{\binom{n}{k}\binom{2n-2k}{n}}\\ &=\sum_{k=0}^\infty (-1)^k[z^k](1+z)^n[u^n](1+u)^{2n-2k}\tag{1}\\ &=[u^n](1+u)^{2n}\sum_{k=0}^\infty\left(-\frac{1}{(1+u)^2}\right)^k[z^k](1+z)^n\tag{2}\\ &=[u^n](1+u)^{2n}\left(1-\frac{1}{(1+u)^2}\right)^n\tag{3}\\ &=[u^n]u^n(2+u)^n\tag{4}\\ &=[u^0](2+u)^n\tag{5}\\ &\color{blue}{=2^n}\tag{6} \end{align*}

Comment:

  • In (1) we apply the coefficient of operator twice. We also set the limit to $\infty$ without changing anything since we are adding zeros only.

  • In (2) we use the linearity of the coefficient of operator and do some rearrangements as preparation for the next step.

  • In (3) we apply the substitution rule of the coefficient of operator with $z:=-\frac{1}{(1+u)^2}$
    \begin{align*} A(u)=\sum_{k=0}^\infty a_k u^k=\sum_{k=0}^\infty u^k [z^k]A(z) \end{align*}

  • In (4) we do some simplifications.

  • In (5) we apply the rule \begin{align*} [u^{p-q}]A(u)=[u^p]u^{q}A(u) \end{align*}

  • In (6) we select the coefficient of $[u^0]$.

Markus Scheuer
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A successful application of Euler´s Finite Difference Theorem

Given $f(x) = \sum_{j=0}^ra_jx^j$, Euler´s Finite Fifference Theorem statest that:

$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}f(k) = (-1)^n \Delta_1^nf(x)\big|_{x=0}=\left\{ \begin{array}{ll} 0, & 0 \leq r<n\\ (-1)^na_nn!,& r = n \end{array} \right. $$


We have \begin{align*} \tag1\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{an-ak}{n} &= \sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{a(n-k)}{n}\\ \tag2&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{(n-k)}\binom{a(n-(n-k))}{n}\\ \tag3&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{n-k}\binom{ak}{n}\\ \tag4&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{k}\binom{ak}{n}\\ \tag5&=(-1)^{n}\sum_{k=0}^{n}(-1)^{-k}\binom{n}{k}\binom{ak}{n}\\ \tag6&=(-1)^{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\binom{ak}{n}\\ \end{align*}

Let $f(k) = \binom{ak}{n}$ where $a$ is any nonzero complex number. The definition of the binomial coefficient implies that $f(k)$ is a polynomial of degree $n$ in $k$, i.e. $\binom{ak}{n} = \sum_{j=0}^na_jk^j$.
The coefficient of $k^n$ is $\frac{a^n}{n!}$ sience

\begin{align*} f(k)=\binom{ak}{n} = \tfrac{(ak)(ak-1)(ak-2)\cdots(ak-n+1)}{n!}=\sum_{j=0}^na_jk^j \end{align*}

By definition

\begin{align*} \tag7(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{ak}{n} &=(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}f(k)\\ \tag8&=(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}\sum_{j=0}^na_jk^j\\ \tag9&=(-1)^n\underbrace{\sum_{j=0}^na_j\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j}_{(-1)^n \Delta_1^nf(x)\big|_{x=0}}=\left\{\begin{array}{ll} 0, & 0\leq j <n\\ a_n n!,& j = n \end{array}\right. \end{align*}


Because here the coefficient of $k^n$ is $\frac{a^n}{n!}$ and $j=n$, implies that

\begin{align*} \tag{10} (-1)^n\sum_{j=n}^na_j\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j&=(-1)^na_n \sum_{k=0}^{n}(-1)^k\binom{n}{k}k^n\\ \tag{11} &=(-1)^n\big[a_n(-1)^nn!\big]\\ \tag{12} &= \frac{a^n}{n!}n! \\ \tag{13} &= a^n\\ &&\Box \end{align*}


In this way we have for $a=2$:

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



Note In $(10)$: When we study Stirling numbers of the second kind, $S(n, k)$, we will discover that $\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j= (-1)^nn!S(n, k),\quad j\leq n$



Bibliographic references:
Gould, H. W. (Oktober 2015). Combinatorial Identities for Stirling Numbers, 68, 69, 70.

Darío A. Gutiérrez
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