# proof that the binomial sum is equal to 1

I'm trying to prove the following identity:

let $k$ and $s$ be positive integers and let $k\ge s\ge 1$ $$\sum_{i=0}^{k-s} (-1)^i{s-1+i \choose s-1}{k \choose s+i} = 1$$

I've tried to use a generating function method to prove this formula, but I don't see how to apply it here just yet. With all the other methods I got stuck. How can it be proven? Thanks in advance.

• You're right, I didn't notice the mistake, thank you for letting me know. it should be correct now – mathew7k5b Mar 25 '16 at 21:06

## 2 Answers

I found it convenient to let $d=k-s$, so that

\begin{align*} \sum_{i=0}^{k-s} (-1)^i{s-1+i \choose s-1}{k \choose s+i}&=\sum_{i=0}^{k-s}(-1)^i\binom{s-1-i}i\binom{k}{k-s-i}\\ &=\sum_{i=0}^{d}(-1)^i\binom{s-1+i}i\binom{s+d}{d-i}\;. \end{align*}

Let

$$f(x)=\sum_{n\ge 0}(-1)^n\binom{s-1+n}nx^n=\sum_{n\ge 0}\binom{s-1+n}n(-x)^n=\frac1{(1+x)^s}$$

and

$$g(x)=\sum_{n\ge 0}\binom{s+d}nx^n=(1+x)^{s+d}\;.$$

Then

$$\sum_{i=0}^{d}(-1)^i\binom{s-1+i}i\binom{s+d}{d-i}$$

is the coefficient of $x^d$ in

$$f(x)g(x)=(1+x)^d\;,$$

which is clearly $1$.

• Thank you very much, it is much clearer to me in this form. I was wondering whether and how we could use this identity to derive a generalized inclusion exclusion principle for the number of elements which are in the intersection of (at least) $s$ sets from $A_1, A_2,..., A_n$? It appears from the sum that what determines whether we add or substract the product of binomial coefficient is whether it has even or odd number of elements and it applies to every subset of s. @Brian – mathew7k5b Mar 27 '16 at 16:32
• @mathew7k4b: You’re very welcome. I’ll have to think about what you’ve asked here. – Brian M. Scott Mar 27 '16 at 17:13

Here is another variation based upon the usage of the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a generating series.

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

Comment:

• In (1) we use the binomial identity \begin{align*} \binom{-n}{k}=\binom{n+k-1}{k}(-1)^k=\binom{n+k-1}{n-1}(-1)^k \end{align*}

• In (2) we apply the coefficient of operator and set the upper limit of the sum to $\infty$ without changing anything, since we add only zeros.

• In (3) we do some rearrangements and use \begin{align*} [z^{n+k}]A(z)=[z^n]z^{-k}A(z) \end{align*}

• In (4) we use the relationship \begin{align*} A(z)=\sum_{j=0}^\infty a_jz^j=\sum_{j=0}^{\infty}\left([u^j]A(u)\right)z^j \end{align*}

• Is there a typo here? $$[u^s] (1+u)^k u^s (1+u)^{-s}$$ While I believe it is useful to feature examples of the formal power series method with this post it would appear very similar to what Brian Scott presented especially the two generating functions. – Marko Riedel Mar 26 '16 at 22:59
• @MarkoRiedel: Typo corrected, thanks. Of course the answers are similar, since they are both based upon generating functions. But, I see the benefit of this answer in the usage of the coefficient of operator, which could be a helpful addon as useful technique. – Markus Scheuer Mar 27 '16 at 5:38