# Value of combinations?

1. Sep 15, 2012

### SMA_01

1. The problem statement, all variables and given/known data

Let n, k be in the set of all positive integers including zero. Define (n k)=n!/k!(n-k)!
Determine the value of $\sum$ (n k) from k=0 to n. Determine the value of $\sum$ (n k) from k=0 to n.

3. The attempt at a solution

I tried evaluating for some n value:

(n 0)= 1

(n 1)= n

(n 2)= (1/2!)n(n-1)

(n 3)= (1/3!)n(n-1)(n-2)

So, generally= 1+n+......+[(1/k!)(n)*.....*(n-(n-k))]+1

Am I on the right track? How can I determine the value?

Any help is appreciated.

2. Sep 15, 2012

### Saitama

I don't think that this problem requires use of Calculus, it can be simply done using the Binomial Theorem.

What is (x+a)n? Write that and try to find the values of x and a by comparing the formula with the series you got i.e. with 1+n+......+[(1/k!)(n)*.....*(n-(n-k))].

http://en.wikipedia.org/wiki/Binomial_theorem

3. Sep 15, 2012

### SMA_01

Pranav-Arora: Thanks. I'm going to have to prove it using induction. Is there anyway other than the binomial theorem to reach the result? I know it will equal 2^n, but I was wondering if you can compute without using the binomial theorem...?

4. Sep 15, 2012

### Saitama

I am not sure if i know of any other method, you should better wait for others members to look at this thread. :)