Prove this sum equal to an expression.

Question

QUESTION:

Prove that for all positive integers $k\leq n$:

$\sum_{i=0}^{k}{n \choose i} (-1)^i = {n-1 \choose k}(-1)^k$

MY THOUGHTS:

INDUCTION : (assuming k even) - We assume:

${n \choose 0} - {n \choose 1} + ... + {n \choose k-2} - {n \choose k-1} = {n-1 \choose k-1}$

We need to show:

${n \choose 0} - {n \choose 1} + ...-{n \choose k-1} + {n \choose k} = {n-1 \choose k-1}$

My issue here is that we can say LHS equals ${n-1 \choose k-1} + {n \choose k}$, which is close, but not quite Pascal's Rule.

Another way: I thought looking to find some bijection that can be applied.

Let's assume n is even. Writing out the left and sum and moving all the odd (negative) terms to the RHS gives (note that the last term on the RHS is simply the original RHS:

${n \choose 0} + {n \choose 2} + ... + {n \choose k} = {n \choose 1} + {n \choose 3} + ... + {n \choose k-1} + {n-1 \choose k}$

I am stuck here, but how I worked out in my mind was that:

$\bigg\{$set of even sized groups of $\leq$ n people$\bigg\}$ --> $\bigg\{$set of odd sized groups of $\leq$ n-1 people$\bigg\}$ + ${n-1 \choose k}$

Then possibly, it would have something to do with adding or removing person n. Let me know your thoughts or suggestions.

Answer 1

There is really nothing to worry about the left hand side only really contains two variables $k$ & $n$. The variable $i$ is referred to as a dummy variable & could have been anything.

Now you are happy that \begin{eqnarray*} \binom{n-1}{k}+\binom{n-1}{k+1}=\binom{n}{k+1}. \end{eqnarray*} Rearrange this to \begin{eqnarray*} \binom{n-1}{k}-\binom{n}{k+1}=\binom{n-1}{k+1} \end{eqnarray*} & now the induction is easy ... \begin{eqnarray*} \sum_{i=0}^{k+1} (-1)^k \binom{n}{k}=(\sum_{i=0}^{k} (-1)^k \binom{n}{k})+ (-1)^{k+1}\binom{n}{k+1} \end{eqnarray*} Now use the inductive hypothesis on the sum in the bracket & then use the equation rearranged above & you are there.

Answer 2

This problem is worked in as equation (5.15) in Concrete Mathematics.
The idea is to use the formula for negating the upper index in a binomial coefficient: $$ \binom{i-r-1}{i} = (-1)^i\binom{r}{i} $$ To prove this, some notation will be useful: Let $r^\underline{i}$, with $i$ a non-negative integer and $r$ any real number, stand for the descending product of $i$ terms $$ r^\underline{0}=1\\ r^\underline{i} = r(r-1)r-2)\cdots (r-i+1)\\ \binom{r}{i} = \frac{r^\underline{i}}{i!} $$

Then by negating each of the $i$ terms in the product that makes up $r^\underline{i}$ we obtain $$ r^\underline{i} = (-1)^i (-r) (1-r) (2-r) \cdots (i-1-r) = (-1)^i (i-1-r)^\underline{i} $$ and so $$ (-1)^i\binom{r}{i} = (-1)^i\frac{r^\underline{i}}{i!}= (-1)^i (-1)^i\frac{(i-1-r)^\underline{i}}{i!} = \binom{i-r-1}{i} $$ We also will use the formula that for non-negative integer $k$ (and integer summation index $i$) $$ \sum_{i\leq k} \binom{r+i}{i} = \binom{r+k+1}{k} $$ This has (at least for integer $r$) a simple combinatorial interpretation:

If I want to choose $b$ items from a collection of $k+(r+1)$ items, I can choose all $r$ from the first $k+r$ items (and not select the last item, or I can choose the last item and $r-1$ out of the first $k+r-1$ items, or I can choose the last two items and $r-1$ out of the first $k+r-2$ items, and so forth.

So we have $$\sum_{i\leq k} (-1)^i\binom{r}{i} =\sum_{i\leq k} \binom{i-r-1}{r} =\sum_{i\leq k} \binom{(-r-1)+i}{i} = \binom{(-r-1)+k+1}{k} = \binom{k-r}{k}\\ \binom{k-(r-1)-1}{k}=(-1)^k \binom{r-1}{k} $$ Replacing $r$ by $n$ we have the desired relation.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\verts{z} < 1}$:

\begin{align} \sum_{i = 0}^{k}{n \choose i}\pars{-1}^{\,i} & = \bracks{z^{k}}\sum_{\ell = 0}^{\infty}z^{\ell} \bracks{\sum_{i = 0}^{\ell}{n \choose i}\pars{-1}^{\,i}} = \bracks{z^{k}}\sum_{i = 0}^{\infty}{n \choose i}\pars{-1}^{\,i} \sum_{\ell = i}^{\infty}z^{\ell} \\[5mm] & = \bracks{z^{k}}\sum_{i = 0}^{\infty}{n \choose i}\pars{-1}^{\,i}\, \pars{z^{i} \over 1 - z} = \bracks{z^{k}}{1 \over 1 - z}\sum_{i = 0}^{\infty}{n \choose i}\pars{-z}^{\,i} \\[5mm] & = \bracks{z^{k}}{1 \over 1 - z}\bracks{\pars{1 - z}^{n}} = \bracks{z^{k}}\pars{1 - z}^{n - 1} = \bracks{z^{k}}\sum_{i = 0}^{\infty}{n - 1 \choose i}\pars{-z}^{\,i} \\[5mm] & = \bbx{\ds{{n - 1 \choose k}\pars{-1}^{\,k}}} \end{align}