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How do I use pascals identity:

$${2n\choose 2k}={2n-1\choose 2k}+{2n-1\choose 2k-1}$$ to prove that

$$\displaystyle\sum_{k=0}^{n}{2n\choose 2k}=\displaystyle\sum_{k=0}^{2n-1}{2n-1\choose k}$$

for every positive integer $n$ ?

darij grinberg
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Tomer Amir
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  • Try writing out Pascal's identity for n = 1, 2, 3, 4 and writing out the equation that you're trying to prove for n = 4 (without simplifying the binomial coefficients). See if you can give a proof in that case. If you're able to do this, you'll see how to do the general case. – Jonah Sinick Oct 26 '12 at 20:30

4 Answers4

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You have $$ \sum_{k=0}^n\binom{2n}{2k}=\sum_{k=0}^n\left(\binom{2n-1}{2k}+\binom{2n-1}{2k-1}\right) $$ Expand the right side. Here are first two and the last terms: $$ \left(\binom{2n-1}{0}+\binom{2n-1}{-1}\right)+\left(\binom{2n-1}{2}+\binom{2n-1}{1}\right)+\cdots+\left(\binom{2n-1}{2n}+\binom{2n-1}{2n-1}\right) $$ Do you see the pattern? Can you express this as a sum? There's only one slightly subtle point.

Rick Decker
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    @Cameron. Where I come from, $\binom{2n-1}{-1}\text{ and }\binom{2n-1}{2n}$ are both defined to be zero. That's a fairly common extension. – Rick Decker Oct 26 '12 at 20:16
  • Never seen that before, but that's good to know (and certainly intuitive). It might be worth noting explicitly in the answer, though, since the binomial coefficient is defined in many different ways depending on the source. – Cameron Buie Oct 26 '12 at 20:48
  • @Cameron. That's why I ended my answer as I did: that was my "subtle point." – Rick Decker Oct 26 '12 at 20:54
  • thanks! i got it! but where can i find the definition of $${2n-1\choose -1}$$ and $${2n-1\choose 2n}$$? i never heard about it, and i would like to read more! – Tomer Amir Oct 27 '12 at 09:40
  • @Tomer. Lots of people do this. For example, Knuth's Concrete Mathematics, Ch.5. In general, $\binom{n}{m}=0$ when $m<0$ and also when $n<m$. In other words, any integer pairs not in Pascal's triangle are zero. – Rick Decker Oct 27 '12 at 16:47
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Other than Pascal's identity, we just notice that the sums on the right are the same because they cover the same binomial coefficients (red=even, green=odd, and blue=both). $$ \begin{align} \sum_{k=0}^n\binom{2n}{2k} &=\sum_{k=0}^n\color{#C00000}{\binom{2n-1}{2k}}+\color{#00A000}{\binom{2n-1}{2k-1}}\\ &=\sum_{k=0}^{2n-1}\color{#0000FF}{\binom{2n-1}{k}} \end{align} $$ Note that $\binom{2n-1}{-1}=\binom{2n-1}{2n}=0$.

robjohn
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  • how come they are equal to 0? – Tomer Amir Oct 27 '12 at 09:47
  • @Tomer: there are three ways to look at this. The first is that $\binom{2n-1}{k}$ is the coefficient of $x^k$ in $(1+x)^{2n-1}$. The coefficients of $x^{2n}$ and $x^{-1}$ are $0$. The second is that, using $\binom{n}{k}=\frac{n!}{(n-k)!k!}$, $k\binom{2n-1}{k}=(2n-k)\binom{2n-1}{k-1}$; set $k=2n$ and $k=0$. The third is using Pascal's identity: $\binom{2n-1}{2n-1}+\binom{2n-1}{2n}=\binom{2n}{2n}$ and $\binom{2n-1}{-1}+\binom{2n-1}{0}=\binom{2n}{0}$. – robjohn Oct 27 '12 at 13:26
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As an aside, I note that the identity

$$\sum_{k=0}^{n}{2n\choose 2k}=\sum_{k=0}^{2n-1}{2n-1\choose k}\tag{1}$$

has an easy combinatorial proof that makes no use of Pascal’s identity. The lefthand side of $(1)$ obviously counts the number of even-sized subsets of $\{1,\dots,2n\}$. The righthand side counts all of the subsets of $\{1,\dots,2n-1\}$. The even-sized subsets of $\{1,\dots,2n\}$ that do not contain $2n$ are obviously in one-to-one correspondence with the even-sized subsets of $\{1,\dots,2n-1\}$, while the even-sized subsets of $\{1,\dots,2n\}$ that do contain $2n$ are in one-to-one correspondence with the odd-sized subsets of $\{1,\dots,2n-1\}$ by the map that throws away $2n$, so the even-sized subsets of $\{1,\dots,2n\}$ are in one-to-one correspondence with the subsets of $\{1,\dots,2n-1\}$, and $(1)$ follows immediately. (And of course both summations are equal to $2^{2n-1}$.)

Another way to say the same thing is to observe that for $n\ge 1$, half of the subsets of $\{1,\dots,n\}$ have even cardinality. (This is easily proved by induction, for instance.) $\{1,\dots,2n\}$ has $2^{2n}$ subsets; half of these, or $2^{2n-1}$, have even cardinality. But that’s the number of all subsets of $\{1,\dots,2n-1\}$.

It should be noted that $(1)$ holds only for $n>0$: if $n=0$, the lefthand side is $1$, and the righthand side is $0$.

Brian M. Scott
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HINT

Write out what you want to prove for $n = 1$ and $n = 2$. Use Pascal's identity. If you are stuck after this, update your question with your work and we'll go from there.

davidlowryduda
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