0

By examination, it's evident that adding every other term in a row of Pascal's triangle gets you $2^{n-1}$. This either follows by induction, properties of Pascal's triangle, the binomial theorem, etc. However, I am wondering if there is a direct combinatorial way to see this.

RobPratt
  • 50,938
Emperor Concerto
  • 1,446
  • 1
    Pascal's triangle is defined inductively, so you will surely need to use some form of induction, hidden or not. – TonyK Jul 17 '21 at 11:21
  • Isn't just a consequence of $n-1 \choose k$ being number of subsets of $k$ elements taken from a total of $n-1$, and all possible subsets from all size are exactly $2^{n-1}$? I think that counts as combinatorial. – AnilCh Jul 17 '21 at 11:37
  • @AnilCh: the OP said "adding every other term". That means $a_0+a_2+a_4+\cdots$ or $a_1+a_3+a_5+\cdots$ – TonyK Jul 18 '21 at 21:18
  • 1
    @TonyK Oh, thank you for clarifying, English is not my native tongue and I didn't know that. – AnilCh Jul 19 '21 at 07:24

1 Answers1

4

When you add every other term of row $n$, you count all the subsets of an $n$-set with odd size, or all the subsets with even size. But there are as many odd-size subsets as there are even-size subsets: if $n$ is odd then the complement is a bijection, if $n$ is even the same result can be obtained by fixing a particular element (remove-the-special-element is a bijection between odd-size subsets with the special element and even size subsets without the special element; add-the-special-element is a bijection between odd-size subsets without the special element and even-size subsets with the special element). Thus the sum is half the total number of subsets of an $n$-set.

Matthew Towers
  • 11,540