# Combinatorics: 1= ((n+k)Cr)*((1/2)^(n+k)) from k = 0 to n

Gold Member
(This is not HW, even though it may look a bit like it.)
Using the notation nCr in the combinatorics meaning,
The sum of ((n+k)Cn)*((1/2)^(n+k)) from k = 0 to n equals one. Why?
(I thought it might use the identity (n+1)Cr = nCr+nC(r-1), but that didn't get me anywhere.)
Thanks in advance for any pointers.

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#### Stephen Tashi

Gold Member
oops, right. Thanks, Stephen Tashi, for spotting the typo. I have edited it.
So, back to the question......

#### homeomorphic

Not sure you've stated it correctly, still. Is this what you wanted?

$1 = 1^n = (\frac{1}{2} + \frac{1}{2} )^n$

$= \Sigma_{k = 0}^n {{n}\choose{k}} \frac{1}{2^n}$ (binomial expansion)

Gold Member
Thanks, homeomorphic, but no: my title stated it incorrectly (I forgot to edit the title, and now it no longer gives me that option), but I have edited the original question, as Stephen Tashi pointed out. The binomial expansion you suggest only gives me
Sum of (nCk)*((1/2)^(k)) from k = 0 to n
Note the differences to
Sum of ((n+k)Cn)*((1/2)^(n+k)) from k = 0 to n
As a random example, I asked Wolfram Alpha to compute this, with n=7, and it was indeed one. This is not a proof, but a reassurance that I wrote it right.

#### micromass

(This is not HW, even though it may look a bit like it.)
And if it looks like homework, then it belongs in the homework forums! I'll move it for you

Gold Member
OK, thanks. Hope to get a response there. As a side note: the question arose as I was looking at Banach's matchbox problem.

Gold Member
reformulated:sum from k=0 to n of (n+k)Ck = 2^n

Or, to put it another way, why is the sum from k=0 to n of ([(n+k)Ck]/2k) equal to 2n?

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#### micromass

Or, to put it another way, why is the sum from k=0 to n of ([(n+k)Ck]/2k) equal to 2n?
I'm confused what the problem actually is. Is this the problem:

$$\sum_{k=0}^n \binom{n+k}{k} \frac{1}{2^k} = 2^n$$

because the OP and the title both say completely different things.

#### micromass

I'm confused what the problem actually is. Is this the problem:

$$\sum_{k=0}^n \binom{n+k}{k} \frac{1}{2^k} = 2^n$$

because the OP and the title both say completely different things.
Or is it

$$\sum_{k=0}^n \binom{n+k}{n} \frac{1}{2^k} = 2^n$$

Gold Member
They are the same thing, since (n+k)Ck = (n+k)C[(n+k)-n]=(n+k)Cn.
I merely restated it in the two fashions, because perhaps one formulation is easier to prove than the other.

#### micromass

They are the same thing, since (n+k)Ck = (n+k)C[(n+k)-n]=(n+k)Cn.
I merely restated it in the two fashions, because perhaps one formulation is easier to prove than the other.
Oh right, haha. Shame on me for missing that... I'm gonna think about this for a while. I assume you have already tried induction?

Gold Member
Thanks for considering it, micromass. Of course I tried induction, but got stuck. This does not mean that induction is not the right way to go; it could very well just mean that I am missing an obvious step in the induction. In the induction step I get
Assume
sum from k=0 to n of ([(n+k)C(k)]/2k) = 2n
Prove
sum from k=0 to n+1 of ([(n+k+1)C(k)]/2k) = 2n+1
Then
sum from k=0 to n+1 of ([(n+k+1)C(k)]/2k) = [sum from k=0 to n of ([(n+k+1)Ck]/2k)]+[(2n+1)C(n+1)/2n+1] and then?
(similarly with the other formulation)

Gold Member
Thanks very much, micromass! That's a path I had not considered. Very elegant.

#### micromass

Thanks very much, micromass! That's a path I had not considered. Very elegant.
Yeah, it's also not entirely correct. You need to do something with the Taylor series

$$\frac{y^n}{(1-y)^{n+1}} = \sum_{k=n}^{+\infty} \binom{k}{n} x^k$$

But then you also need some formula for

$$\sum_{k=2n}^{+\infty} \binom{k}{n}x^k$$

which I haven't found yet.

Gold Member
Thanks again, micromass; at least for the moment it gives me another line of thought to pursue. Although this is in the "HW & HW-similar" section, it is not HW and so there is no "due date", so if you come up with something anytime in the future, I would be grateful to hear about it.

#### micromass

OK, a highly nontrivial way is this:

We are interested in

$$x_a = \sum_{k=a}^{+\infty} \binom{k}{n} \frac{1}{2^k}$$

If we found this, then your sum is simply $x_{n} - x_{2n+1}$. Now, it is easily seen that

$$x_a = \binom{a}{n} \frac{1}{2^a} {}_2F_1(1,a+1;a-n+1;1/2)$$

Where ${}_2F_1$ is the hypergeometric function (it's really easy to see, it just follows from the definition, I might have made some computation errors tho). So we want to calculate

$${}_2F_1(1,n+1;1;1/2)~\text{and}~{}_2F_1(1,2n+2;n+3;1/2)$$

The former can be calculated by applying a Euler transformation and yields $2^{n+1}$: http://en.wikipedia.org/wiki/Hypergeometric_function#Fractional_linear_transformations

The first one, I can't find at the moment.

Gold Member
Thanks, micromass. As you wrote, non-trivial, but it gives me some new techniques to get acquainted with.

#### micromass

The first one, I can't find at the moment.
The second one $${}_2F_1(1,2n+2;n+3;1/2)$$ can be calculated by applying Gauss' contiguous relations. Specifically, I'm applying

$$\frac{(c-a){}_2F_1(a-1,b;c;z) + (a -c + bz){}_2F_1(a,b;c;z)}{1-z} = (c-1)({}_2F_1(a,b;c-1;z) - {}_2F_1(a,b;c;z))$$

This means that we simply need to find ${}_2F_1(0,2n+2;n+3;1/2)$ which is easily done by applying the definition. We also need to find ${}_2F_1(1,2n+2;n+2;1/2)$. But this can be done by Gauss' second summation theorem: http://en.wikipedia.org/wiki/Hypergeometric_function#Values_at_z.C2.A0.3D.C2.A01.2F2

#### micromass

Well, this works, but an elementary approach still eludes me.

#### micromass

OK, but that solves it right? I don't get why you say "special case"? I'll write it here for future reference:

You need to find

$$\sum_{k=0}^n \binom{n+k}{n} \frac{1}{2^{n+k}}$$

Multiply by $4^n/4^n$ to get

$$\frac{1}{4^n} \sum_{k=0}^n \binom{n+k}{n} 2^{n-k}$$

Apply the relation $2^a = \sum_{j=0}^a \binom{a}{j}$ to get

$$\frac{1}{4^n} \sum_{k=0}^n \binom{n+k}{n} \sum_{j=0}^{n-k} \binom{n-k}{j}$$

Switching around the sums and allowing the convention that if $\binom{a}{b}=0$ if $b>a$, we get

$$\frac{1}{4^n} \sum_{j=0}^n \sum_{k=0}^n \binom{n-k}{j}\binom{n+k}{n}$$

the relation in Knuth is

$$\sum_{k=0}^l \binom{l-k}{m}\binom{q+k}{n} = \binom{l+q+1}{m+n+1}$$

Thus we get

$$\frac{1}{4^n} \sum_{j=0}^n \binom{2n+1}{j+n+1}$$

Change variables $l=j+n+1$:

$$\frac{1}{4^n} \sum_{l=n+1}^{2n+1} \binom{2n+1}{l}$$

Applying that $\binom{2n+1}{l} = \binom{2n+1}{2n+1-l}$, we see that

$$\frac{1}{2^{2n-1}} \sum_{l=0}^{2n+1} \binom{2n+1}{l}$$

Again applying $2^a = \sum_{j=0}^a \binom{a}{j}$, we see that this is equal to

$$\frac{1}{2^{2n+1}}2^{2n+1} = 1$$

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Gold Member
Thanks again, micromass. Yes, the phrase "a special case" was misplaced; I was being overly careful before I had seen the reference in Knuth et al. (I have in fact not yet obtained a copy, so I still need to look at it.) Apparently, as you say, this now solves it, and your exposition is nice and clear -- except that I think you left something out in the line
"Applying that $\binom{2n+1}{l} = \binom{2n+1}{l}$, we see that..."

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#### micromass

Thanks again, micromass. Yes, the phrase "a special case" was misplaced; I was being overly careful before I had seen the reference in Knuth et al. (I have in fact only got a copy of it a few minutes ago, so I still need to look at it.) Apparently, as you say, this now solves it, and your exposition is nice and clear -- except that I think you left something out in the line
"Applying that $\binom{2n+1}{l} = \binom{2n+1}{l}$, we see that..."
Thank you, that was a typo. Well, all that remains is proving the relation in Knuth, which is not done there. I think an induction proof works, but I don't like that. Consider the function

$$f_n(x) = \frac{x^n}{(1-x)^{n+1}}$$

We know the binomial series

$$\frac{1}{(1-x)^{n+1}} = \sum_{k=0}^{+\infty} \binom{k+n}{n} x^k$$

Thus

$$f_n(x) = \sum_{k=0}^{+\infty} \binom{k}{n} x^k$$

with the usual convention that $\binom{k}{n} = 0$ if $n>k$

Thus $xf_n(x)f_m(x) = f_{n+m+1}(x)$. By the Cauchy product of series, we then see that

$$\sum_{k=-1}^{+\infty} \binom{k+1}{n+m+1} x^k = \sum_{k=0}^{+\infty} x^k \sum_{j=0}^k \binom{j}{n}\binom{k-j}{m}$$

Thus we see the identity (with additionally substituting $t=k$

$$\sum_{j=0}^k \binom{j}{n}\binom{t-j}{m} = \binom{t+1}{n+m+1}$$

Substituting $j=l-k$, we get

$$\sum_{k=0}^{l-k} \binom{l-k}{n}\binom{t-l+k}{m} = \binom{t+1}{n+m+1}$$

The result now follows after substiting $q=t-l$.

Gold Member
Excellent! Thanks so very much. Yes, I finally got hold of Knuth and was surprised not to find a derivation for the the principle in question. So your explanation definitely fills a gap, completing the answer to my question. I am very grateful.

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