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Thread: Even vs. Odd in a combinatoric proof

  1. #1
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    Even vs. Odd in a combinatoric proof

    I was given a problem to simplify the following:
    $\displaystyle \sum_{k \geq 0} \binom{n}{2k} $
    where:
    $\displaystyle \binom{n}{j} = 0 $ whenever j > n.
    It's easy to see that the problem is equivalent to:
    $\displaystyle \sum_{k = 0}^n \binom{n}{2k} $

    Now I can see that for odd values of n, this expands to:
    $\displaystyle \binom{n}{0} + \binom{n}{2} + ... + \binom{n}{n-1} $
    and since:
    $\displaystyle \binom{n}{n-1} = \binom{n}{1}, \binom{n}{n-3} = \binom{n}{3}, ... $
    we can use the last n/2 terms to "fill in the gaps", simplifying it to be:
    $\displaystyle \frac{\sum_{k=0}^n \binom{n}{k}}{2} = \frac{2^n}{2} = 2^n-1 $

    That's easy. But I don't know why the proof still works for even numbers of n. You no longer have the same correspondence (n choose 0 no longer has the equivalent n choose n-1 as a term):
    $\displaystyle \binom{n}{0} + \binom{n}{2} + ... + \binom{n}{n-2} + \binom{n}{n} $
    So it can't "fill in the gaps" in the same way.

    Can anyone figure this out? I've been at this for hours and the explanations I'm being given by the texts are just crap.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by RespeckKnuckles View Post
    I was given a problem to simplify the following:
    $\displaystyle \sum_{k \geq 0} \binom{n}{2k} $
    where:
    $\displaystyle \binom{n}{j} = 0 $ whenever j > n.
    It's easy to see that the problem is equivalent to:
    $\displaystyle \sum_{k = 0}^n \binom{n}{2k} $

    Now I can see that for odd values of n, this expands to:
    $\displaystyle \binom{n}{0} + \binom{n}{2} + ... + \binom{n}{n-1} $
    and since:
    $\displaystyle \binom{n}{n-1} = \binom{n}{1}, \binom{n}{n-3} = \binom{n}{3}, ... $
    we can use the last n/2 terms to "fill in the gaps", simplifying it to be:
    $\displaystyle \frac{\sum_{k=0}^n \binom{n}{k}}{2} = \frac{2^n}{2} = 2^n-1 $

    That's easy. But I don't know why the proof still works for even numbers of n. You no longer have the same correspondence (n choose 0 no longer has the equivalent n choose n-1 as a term):
    $\displaystyle \binom{n}{0} + \binom{n}{2} + ... + \binom{n}{n-2} + \binom{n}{n} $
    So it can't "fill in the gaps" in the same way.

    Can anyone figure this out? I've been at this for hours and the explanations I'm being given by the texts are just crap.
    Here is a combinatorial proof. Clearly $\displaystyle \displaystyle \sum_{k=0}^{n}{n\choose {2k}}$ counts the number of subsets of $\displaystyle [n]$ with
    an even number

    of elements. Call the set of subsets with an odd number elements of $\displaystyle [n]$ $\displaystyle O_n$ and the set of subsets with an

    even number elements $\displaystyle E_n$. We claim that $\displaystyle \#\left(E_n\right)=\#\left(O_n\right)$ for every $\displaystyle n\in\mathbb{N}$. The result is clearly true

    for $\displaystyle n=1$. Suppose it's true for $\displaystyle [n]$ and note that $\displaystyle [n+1]$ can be partitioned into the subsets of $\displaystyle [n+1]$

    which contain $\displaystyle n+1$ and those that don't, call this partition $\displaystyle \{B_1,B_2\}$ (with $\displaystyle B_1$ being the collection

    of subsets not containing $\displaystyle n+1$) it's evident that $\displaystyle O_{n+1}$ is equal to the number of subsets with an odd number

    of elements of $\displaystyle B_1$ plus the number of subsets of $\displaystyle B_2$ with an odd number of elements and similarly for $\displaystyle E_{n+1}$.

    But, it's clear that the number of subsets of $\displaystyle B_1$ with an odd number of elements is $\displaystyle O_{n}$ and the number of

    subsets of $\displaystyle B_1$ with an even number of elements is $\displaystyle E_n$. But, a little thought shows that the number of

    subsets with an even number of elements in $\displaystyle B_2$ is $\displaystyle O_n$ and the number of subsets with an odd number of

    elements is $\displaystyle E_n$. Thus, $\displaystyle O_{n+1}=O_n+E_n$ and $\displaystyle E_{n+1}=O_n+E_n$ from where the conclusion follows by

    induction. Thus,


    $\displaystyle \displaystyle \begin{aligned}\sum_{k=0}^{n}{n\choose 2k} &=\frac{1}{2}\left(\#(E_n)+\#(E_n)\right)\\ &= \frac{1}{2}\left(\#(E_n)+\#(O_n)\right)\\ &= \frac{1}{2}2^n\\ &= 2^{n-1}\end{aligned}$




    Remark: Incidentally we've proven that $\displaystyle \#2([n])=\#([n+1])$ and since $\displaystyle \#(2^{\varnothing})=1$ we may conclude that

    $\displaystyle \#([n])=2^n$
    Last edited by Drexel28; Jan 2nd 2011 at 12:42 AM.
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  3. #3
    Senior Member BAdhi's Avatar
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    you've taken $\displaystyle \sum_{k\geq0} \rightarrow \sum_{k=0}^n$ but doesn't it has to be $\displaystyle \sum_{k\geq0} \rightarrow \sum_{k=0}^{n/2}$ that is because when$\displaystyle 2k>n$,$\displaystyle \binom{n}{2k}=0 $so the largest value that k can have is $\displaystyle \frac{n}{2}$. if this is correct, since k is an integer, n has to even.
    Last edited by BAdhi; Jan 2nd 2011 at 12:56 AM. Reason: latex errors
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  4. #4
    Senior Member BAdhi's Avatar
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    correction to the previous post

    forget what i've mentioned in my previous post it's all wrong and really sorry abount it.(because n is a given value it can either be odd or even)

    but check out this method. Hope it's not wrong too.

    $\displaystyle (a+b)^n=\sum_{r=0}^n\binom{n}{r}a^rb^{n-r}$

    if n is odd

    if a=1,b=-1

    $\displaystyle 0=\sum_{r=0}^n\binom{n}{2r}-\sum_{r=0}^n\binom{n}{2r+1}$ ___ (1)

    if a=b=1;

    $\displaystyle 2^n=\sum_{r=0}^n\binom{n}{2r}-\sum_{r=0}^n\binom{n}{2r+1}$___(2)

    by (1)+(2) i think we'll get the result for odd n

    when n is even, same 2 equations could be achieved.
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