How can I compute $\sum\limits_{k = 1}^n \binom{n - 1}{k - 1}$?

Question

I know what $n \choose k$ equals, but I don't see how that would help me solve the sum of $n - 1 \choose k - 1$ from $k = 1$ to $n$. Is there any special trick I should know?

See also: http://math.stackexchange.com/questions/18690/algebraic-proof-that-sum-limits-i-0n-binomni-2n and the posts shown there among linked questions. (The answers has already shown you how to modify your problem to the other sum.) — Martin Sleziak, Feb 09 '15 at 20:31

score 4 · Answer 1 · edited Feb 09 '15 at 19:50

4

Hint: Binomial Expansion of $(1+1)^{n-1} = \sum_{k=1}^{n-1}{n-1 \choose k-1}1^k1^{n-k}$ will lead to the sum you have asked for.

edited Feb 09 '15 at 19:50

DanielV

23,556

answered Feb 09 '15 at 19:47

kryomaxim

2,882

score 4 · Answer 2 · answered Feb 09 '15 at 19:48

4

Change the index by setting $j = k - 1$. Then $$\sum_{k = 1}^n \binom{n-1}{k-1} = \sum_{j = 0}^{n-1}\binom{n-1}{j} = (1 + 1)^{n-1} = 2^{n-1}.$$

answered Feb 09 '15 at 19:48

kobe

41,901

score 3 · Accepted Answer · answered Feb 09 '15 at 19:51

Here's a combinatorial way to approach the problem if you understand what the binomial coefficient ${{n-1} \choose {k-1}}$ means. It's the number of ways you can choose a (unordered) subset of size $k-1$ from a (unordered) set of $n-1$ elements. So if your sum is from $k=1$ to $k = n$, then you are adding up all the ways you can choose an arbitrary unordered subset of any size, from 0 up to $n-1$. This is $2^{n-1}$, because for each of the $n-1$ elements in your set, you have two choices: Either the element is in the subset or not.

How can I compute $\sum\limits_{k = 1}^n \binom{n - 1}{k - 1}$?

3 Answers3