An alternating sum with binomial coefficients

Question

How to calculate this sum $$ \sum_{k=0}^{n-1}(-1)^k{n \choose k} {3n-k-1 \choose 2n -k } $$ without complex integral technique? Or how to calculate asymptotic nature this sum without calculation of this sum?

Answer 1

I like combinatorial solutions, and the form of the sum

$$\sum_{k=0}^{n-1}(-1)^k\binom{n}k\binom{3n-k-1}{2n-k}$$

immediately suggests that it could result from an inclusion-exclusion calculation, though it is missing what would normally be the final term,

$$(-1)^n\binom{2n-1}n\;.$$

It turns out to be convenient to make use of the fact that

$$\binom{3n-k-1}{2n-k}=\binom{3n-k-1}{n-1}$$

to rewrite the summation as

$$\sum_{k=0}^{n-1}(-1)^k\binom{n}k\binom{3n-k-1}{n-1}\;.$$

Suppose that I want to count the $(n-1)$-element subsets of $[3n-1]\setminus[n]$, where as usual $[m]=\{1,\ldots,m\}$ for any $m\in\Bbb Z^+$. For each $k\in[n]$ let $\mathscr{A}_k$ be the family of $(n-1)$-element subsets of $[3n-1]$ that do not contain $k$. It’s not hard to see that if $\varnothing\ne I\subseteq[n]$, then

$$\left|\bigcap_{k\in I}\mathscr{A}_k\right|=\binom{3n-1-|I|}{n-1}\;,$$

so

$$\left|\bigcup_{k\in[n]}\mathscr{A}_k\right|=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\left|\bigcap_{k\in I}\mathscr{A}_k\right|=\sum_{k=1}^n(-1)^{k+1}\binom{n}k\binom{3n-1-k}{n-1}\;.$$

This is the number of $(n-1)$-element subsets $S$ of $[3n-1]$ such that $[n]\nsubseteq S$, so

$$\begin{align*} \binom{3n-1}{n-1}-\sum_{k=1}^n(-1)^{k+1}\binom{n}k\binom{3n-1-k}{n-1}&=\binom{3n-1}{n-1}+\sum_{k=1}^n(-1)^k\binom{n}k\binom{3n-1-k}{n-1}\\ &=\sum_{k=0}^n(-1)^k\binom{n}k\binom{3n-1-k}{n-1} \end{align*}$$

is the number of $(n-1)$-element subsets $S$ of $[3n-1]$ such that $[n]\subseteq S$. This is obviously $0$, so

$$\sum_{k=0}^n(-1)^k\binom{n}k\binom{3n-1-k}{n-1}=0\;,$$

and

$$\begin{align*} \sum_{k=0}^{n-1}(-1)^k\binom{n}k\binom{3n-1-k}{n-1}&=-(-1)^n\binom{2n-1}{n-1}\\ &=(-1)^{n+1}\binom{2n-1}{n-1}\\ &=(-1)^{n+1}\binom{2n-1}n\;. \end{align*}$$

Answer 2

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

We obtain for $n\geq 1$: \begin{align*} \sum_{k=0}^{n-1}&(-1)^k\binom{n}{k}\binom{3n-k-1}{2n-k}\\ &=\sum_{k=0}^{n}\binom{n}{k}\binom{-n}{2n-k}-\binom{-n}{n}\tag{1}\\ &=\sum_{k=0}^\infty[z^k](1+z)^n[u^{2n-k}](1+u)^{-n}+\color{blue}{(-1)^{n-1}\binom{2n-1}{n}}\tag{2}\\ &=[u^{2n}](1+u)^{-n}\sum_{k=0}^\infty u^k[z^k](1+z)^n+(-1)^{n-1}\binom{2n-1}{n}\tag{3}\\ &=[u^{2n}]1+(-1)^{n-1}\binom{2n-1}{n}\tag{4}\\ &=(-1)^{n-1}\binom{2n-1}{n} \end{align*}

Comment:

• In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^k$. We also add the term with index $k=n$ to the sum and subtract $\binom{-n}{n}$ accordingly.

• In (2) we apply the coefficient of operator twice and use again the binomial identity as in (1). We also set the upper limit of the sum to $\infty$ without changing anything since we are adding zeros only.

  In fact we have isolated the result (blue) and show the rest is equal to zero.

• In (3) we use the linearity of the coefficient of operator and apply the rule $$[z^{p-q}]A(z)=[z^p]z^qA(z)$$

• In (4) we use the substitution rule with $z:=u$ \begin{align*} A(u)=\sum_{k=0}^\infty a_k u^k=\sum_{k=0}^\infty u^k [z^k]A(z) \end{align*}

  do some simplifications and observe the coefficient of $[u^{2n}]1=0$.

Answer 3

$(-1)^k\binom{n}{k}$ is the coefficient of $x^k$ in $(1-x)^n$.
$\binom{3n-k-1}{2n-k}=\binom{3n-k-1}{n-1}$ is the coefficient of $x^{2n-k}$ in $$ \sum_{h\geq 0}\binom{n-1+h}{n-1}x^h = \frac{1}{(1-x)^n} $$ It follows that for any $n>0$ $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{3n-k-1}{n-1}=[x^{2n}]\frac{(1-x)^n}{(1-x)^n} = 0 $$ and: $$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{3n-k-1}{n-1}= \color{red}{-(-1)^{n}\binom{n}{n}\binom{2n-1}{n-1}}. $$ Asymptotics can be derived from $\binom{2n}{n}\approx\frac{4^n}{\sqrt{\pi n}}.$

Answer 4

Ok here we go:

$$ s(k,n)=(-1)^k\binom{n}{k}\binom{3n-k-1}{2n-k}= (-1)^k\binom{n}{k}\binom{-(-n)+ 2n-k-1}{2n-k}=\binom{n}{k}\binom{-n}{2n-k} $$

See here

Furthermore in the form above it is clear that

$$ \sum_{n=0}^{2n}s(k,n)=0 \quad \color{red}{(1)} $$

by Vandermonde's identiy.

Furthermore $$s(k,n)=0 \quad \text{for} \quad k>n \quad \color{blue}{(2)} $$ by definition of the Binomial coefficent.

Combining $\color{red}{(1)}$ and $\color{blue}{(2)}$ yields

$$ \sum_{n=0}^{n-1}s(k,n)=-s(n,n) $$

or

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

Answer 5

$$ \begin{align} \sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{3n-k-1}{2n-k} &=(-1)^{n+1}\binom{2n-1}{n}+\sum_{k=0}^n\binom{n}{k}\binom{-n}{2n-k}\tag{1}\\ &=(-1)^{n+1}\binom{2n-1}{n}+\binom{0}{2n}\tag{2}\\[4pt] &=(-1)^{n+1}\binom{2n-1}{n}+[n=0]\tag{3} \end{align} $$ Explanation:
$(1)$: add and subtract the $k=n$ term
$\phantom{(1)\text{:}}$ use negative binomial coefficients to get $(-1)^k\binom{3n-k-1}{2n-k}=\binom{-n}{2n-k}$
$(2)$: Vandermonde's Identity
$(3)$: $\binom{0}{2n}=[n=0]$ using Iverson Brackets

Answer 6

I wrote a python script to compute the first few values of the sequence:

$n$: $f(n)$

2: -3

3: 10

4: -35

5: 126

6: -462

7: 1716

8: -6435

9: 24310

10: -92378

11: 352716

12: -1352078

13: 5200300

14: -20058300

15: 77558760

16: -300540195

17: 1166803110

18: -4537567650

19: 17672631900

I inputted the first 10 of those values into the OEIS and found that your sum appears to be:

$$(-1)^{n-1}\binom{2n-1}{n} = \sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{3n-k-1}{2n-k}$$

The LHS below should be much easier to analyze asymptotically (use Stirling's approximation). The two sequences match on at least the first 20 or so terms (all I checked). You may want to prove (induction perhaps) that they are the same--unless you believe/are convinced enough that 20+ term match implies they are the same already.