Proof that sum of binomial coefficients equals 2^n

[bennyska]

Homework Statement

(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

Homework Equations

binomial theorem

The Attempt at a Solution

proof: sum(k=0 to n of (n,k)) = sum(k=0 to n of (n,k))*1^k*1^n-k.
by the binomial theorem, (x + y)ⁿ = sum(k=0 to n of (n,k))*x^ky^n-k, so letting x, y = 1, then (1 + 1)ⁿ = 2ⁿ = sum(k=0 to n of (n,k))*1^k*1^n-k = sum(k=0 to n of (n,k)).

 

[rock.freak667]

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

 

[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.

 

[rock.freak667]

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

 

[Gib Z]

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.

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