# Thread: Sum of C(n,r) proof

1. ## Sum of C(n,r) proof

Hi .
Can Anyone help to proof $\sum C(n,r)=2^n$ without using

sets .

I am Wondering how to Solve this Series?

$\sum_{r=0}^{n} \frac{n!}{r!(n-r)!}$

............

I Know :

$\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} = n!\sum_{r=0}^{n} \frac{1}{r!(n-r)!}$

2. Originally Posted by parkhid
Hi .
Can Anyone help to proof $\sum C(n,r)=2^n$ without using

sets .

I am Wondering how to Solve this Series?

$\sum_{r=0}^{n} \frac{n!}{r!(n-r)!}$

............

I Know :

$\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} = n!\sum_{r=0}^{n} \frac{1}{r!(n-r)!}$

We know (I hope...) that $(a,b)^n=\sum\limits_{k=0}^n \binom{n}{k} a^{n-k}b^k$ , so :

$\sum\limits_{k=0}^n \binom{n}{k}=\sum\limits_{k=0}^n \binom{n}{k}1^{n-k}\cdot 1^k=$ ...

Tonio

3. The Binomial Expansion is defined as:

$(x+a)^{n}=\sum_{k=0}^{n}{n \choose k}.x^{n-k}.a^{k} ,\forall x,a$ and $n \in \mathbb{Z}_{+}$

Now, if x=a=1 , We have:

$2^{n}=\sum_{k=0}^{n}{n \choose k}$

What is the sum...

4. ## NO

I will Know this way. i want to proof this by series and differential relations.

5. You can prove that $\sum_{k=0}^n {n \choose k} = 2^n$ by induction on n.

If n=1, then ${1 \choose 0} + {1 \choose 1} = 2$, so it is true for n=1.

Now using the fact that ${n \choose k} + {n \choose {k+1}} = {{n+1} \choose {k+1}}$, and assuming that $\sum_{k=0}^n {n \choose k} = 2^n$, we have that

$\sum_{k=0}^{n+1} {{n+1} \choose k} = \sum_{k=1}^{n+1} {n \choose {k-1}} + \sum_{k=0}^n {n \choose k}$

$\sum_{k=0}^{n+1} {{n+1} \choose k} = \sum_{k=0}^{n} {n \choose k} + \sum_{k=0}^n {n \choose k}$

$\sum_{k=0}^{n+1} {{n+1} \choose k} = 2 \sum_{k=0}^{n} {n \choose k} = 2 \cdot 2^n = 2^{n+1}$