How do I show $(a+b)^K=\sum _{n=0}^K \binom{K}{n} b^n a^{K-n} $ without using recurrence? We already know that we can represent this binomial as the following:
$$(a+b)^K=\sum _{n=0}^K \binom{K}{n} b^n a^{K-n};$$
where $\binom{K}{n} = \frac{K!}{n! (K-n)!}$
My question here is  :How do I show this :$(a+b)^K=\sum _{n=0}^K \binom{K}{n} b^n a^{K-n};  $ is true without using recurrence demonstration?
Note: I know only recurrence method.
Thank you for any help
 A: Calculus proof:
It suffices to show it's true for $(1+x)^K$ since you can factor out $a$ and let $x=b/a$. Then
\begin{align*}
[x^n](1+x)^K&=\frac{1}{n!}\left.\frac{d^n}{dx^n}(1+x)^K\right|_{x=0}\\
&=\frac{1}{n!}K(K-1)\cdots (K-n+1)=\binom{K}{n}.
\end{align*}
So the coefficient infront of $x^n$ is $\binom{K}{n}$. Thus:
$$(a+b)^K=a^K(1+b/a)^K=a^K\sum_{n=0}^K\binom{K}{n}(b/a)^n=\sum_{n=0}^K\binom{K}{n}b^na^{K-n}.$$
Note that the sum has to start from $n=0$ since that's the lowest power of $x$ and also terminates at $n=K$ because that's the highest power (and also the binomial coefficient vanishes).
Combinatorics proof:
The coefficient $a^nb^{N-k}$ occurs as my times as one can pick sequences of length $K$ with exactly $n$ $a$'s (without regard to order) and $n-K$ $b$'s. from the set $\{a,b\}^K$. Equivalently, you are picking $n$ team members from a team of size $K$ with to form team $a$, the remaining ones form team $b$.  This is exactly $\binom{K}{n}=\binom{K}{K-n}$. 
A: Really, $(a+b)^K=\displaystyle\sum_{n=0}^K\dbinom{K}na^n b^{K-n}$ is the definition of the binomial coefficients.
Then considering the equality $\bigl((1+x)^K\bigr)'=K(1+x)^{K-1},\enspace K\ge 1$, and the developments of both sides, you can establish the relation:
$$\binom Kn=\frac Kn\binom{K-1}{n-1},$$ 
from which you deduce by a trivial induction (or with a telescoping product):
$$\binom Kn=\frac{K(K-1)\dots(K-n+1)}{n!}=\frac{K!}{n!\,(K-n)!}$$
