# Homework Help: Please check my proof of sum of n choose k = 2^n

1. Jun 18, 2010

### bennyska

1. The problem statement, all variables and given/known data
(here (n,k) reads n choose k)(and again, please excuse that i don't use latex)
claim: (n,0) + (n,1) + (n,2) + ... (n,n) = 2n

2. Relevant equations

binomial theorem

3. The attempt at a solution
proof: sum(k=0 to n of (n,k)) = sum(k=0 to n of (n,k))*1k*1n-k.
by the binomial theorem, (x + y)n = sum(k=0 to n of (n,k))*xkyn-k, so letting x, y = 1, then (1 + 1)n = 2n = sum(k=0 to n of (n,k))*1k*1n-k = sum(k=0 to n of (n,k)).

2. Jun 18, 2010

### rock.freak667

Looks correct, I see nothing wrong with it. (assuming you weren't given a set method such as mathematical induction to use)

3. Jun 19, 2010

### bennyska

no. but looking back on it, should i have been a little more rigorous? like being explicitly clear that for the binomial theorem, n has to be a positive integer. it's kind of implied, i guess.

4. Jun 19, 2010

### rock.freak667

The actual proof itself looks valid, you can state the conditions before it though.

5. Jun 19, 2010

### Gib Z

The theorem in fact remains true for any complex number n if you use generalized binomial coefficients and binomial series. See http://en.wikipedia.org/wiki/Binomial_series