Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The question is whether the below is true.

$$\sum _{k=0}^{s-1} \binom{n}{k}=\sum _{k=1}^s 2^{k-1} \binom{n-k}{s-k}$$

Mathematica can simplify as follows, but it fails to Reduce[] or Solve[].

$$2^n=\binom{n}{s} \, _2F_1(1,s-n;s+1;-1)+\binom{n-1}{s-1} \, _2F_1(1,1-s;1-n;2)$$

share|improve this question
4  
I guess the easiest way is to check that both sides satisfy the same recurrence relation (as sequences in s), i.e., $-(s + 2)f(s + 2) + (n + 1)f(s + 1) + (s - n + 1)f(s)= 0$. –  Martin Rubey Jun 28 at 15:58
    
MJK, you now have two answers to your question. If you find one or the other satisfactory, then you should consider accepting one of them as a solution: mathoverflow.net/help/accepted-answer –  Todd Trimble Jun 30 at 10:43

2 Answers 2

up vote 22 down vote accepted

A slightly less computational method is to note that both sides of the identity count the number of subsets of $\{1,\dots,n\}$ with fewer than $s$ elements. This is obvious for the left hand side. It's true for the right hand side because $2^{k-1}\pmatrix{n-k\\s-k}$ is the number of such subsets $S$ for which $k$ is minimal such that $|S\cup\{1,\dots,k\}|\geq s$, since such a subset $S$ is the union of an arbitrary subset of $\{1,\dots,k-1\}$ and a subset of size $s-k$ of $\{k+1,\dots,n\}$.

share|improve this answer
    
@MarcvanLeeuwen: $S$ does not contain $k$. –  Timothy Chow Jun 29 at 13:51
    
@TimothyChow: Right. I somehow read that union sign as an intersection though admittedly that makes no sense. I'll leave my silly comment for now, but I'll make it self-destruct in some time. –  Marc van Leeuwen Jun 29 at 14:22

One way is by induction on $n$. We have a series of equations where the first equation is Pascal's triangle identity and the third uses the inductive hypothesis, and the rest is basically by re-indexing sums:

$$\begin{array}{lll} \sum_{k=0}^{s-1} \binom{n}{k} & = & \sum_{k=0}^{s-1} \binom{n-1}{k-1} + \sum_{k=0}^{s-1} \binom{n-1}{k} \\ & = & \binom{n-1}{s-1} + 2\sum_{k=0}^{s-2}\binom{n-1}{k} \\ & = & \binom{n-1}{s-1} + \sum_{k=1}^{s-1} 2^k\binom{n-1-k}{s-1-k} \\ & = & \binom{n-1}{s-1} + \sum_{k=2}^s 2^{k-1}\binom{n-k}{s-k} \\ & = & \sum_{k=1}^s 2^{k-1} \binom{n-k}{s-k}. \end{array}$$

(Not sure this is quite considered "research level", but since the identity of the OP is so cute, I couldn't resist.)

share|improve this answer
5  
I spent a puzzled couple of minutes wondering what was so cute about who the OP is, until I realized that's not what you meant by his/her identity ... –  Jeremy Rickard Jun 28 at 19:57
    
@JeremyRickard lol... I use OP interchangeably for "Original Poster" and "Original Post". Nice bijective proof by the way (although it took me probably about as long puzzling through it as my remark took you). –  Todd Trimble Jun 28 at 20:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.