Prove a summation equals to one How to prove that
$$
S = \sum_{k=0}^{K-1} \binom{k+K-1}{K-1} \frac{a^k b^K + a^K b^k}{(a+b)^{k+K}} \mathop{=}\limits^{?} 1,
$$
For $K=2$, I have S = $\frac{a^2+b^2}{(a+b)^2} + \frac{2ab(a+b)}{(a+b)^3}$, which can be boiled down as $S = \frac{(a+b)(a^2+b^2+2ab)}{(a+b)^3} = 1$. 
Also, WolframAlpha gave the result
$$
S = 2 - \binom{2K-1}{K-1} \frac{(xy)^K}{(x+y)^{2K}} \left\{ {}_2F_1\left(1,2K;K+1;\frac{a}{a+b} \right) + {}_2F_1\left(1,2K;K+1;\frac{b}{a+b} \right) \right\},
$$
but $S = 1$ only when $0 < \frac{a}{a+b} < 1$ and $0 < \frac{b}{a+b} < 1$. 
 A: Write $x={a\over a+b},\ J=K-1$, and 
$$f(x,J) = (1-x)^{J+1}\sum_{k=0}^J{J+k\choose k}x^k$$ Then we want to show $$f(x,J)+f(1-x,J)=1,\ J=0,1,2,\dots\tag{1}$$
We proceed by induction on $J$.  When $J=0$, $(1)$ says $x+(1-x)=1.$
Suppose that $J>0$ and that $(1)$ is true for $J-1$.  Then
$$\begin{align}
f(x,J)&=(1-x)^{J+1}\sum_{k=0}^J{J-1+k\choose k}x^k+(1-x)^{J+1}\sum_{k=0}^J{J-1+k\choose k-1}x^k\tag{2}
\end{align}$$
The first term on the right of $(2)$ is $$
(1-x)^{J+1}\sum_{k=0}^{J-1}{J-1+k\choose k}x^k+(1-x)^{J+1}{2J-1\choose J}x^J$$ or
$$(1-x)f(x,J-1)+(1-x)^{J+1}x^J{2J-1\choose J}\tag{3}$$ 
The second term on the right of $(2)$ is 
$$\begin{align}
(1-x)^{J+1}\sum_{k=1}^J {J-1+k\choose k-1}x^k&=(1-x)^{J+1}\sum_{k=0}^{J-1}{J+k\choose k}x^{k+1}\\
&=xf(x,J)-(1-x)^{J+1}{2J\choose J}x^{J+1}\tag{4}
\end{align}$$
Now $(2),(3),\text{ and }(4)$ give 
$
(1-x)f(x,J)=(1-x)f(x,J-1)+(1-x)^{J+1}x^J{2J-1\choose J}-(1-x)^{J+1}{2J\choose J}x^{J+1}\tag{6}
$
By symmetry, 
$
xf(1-x,J)=xf(1-x,J-1)+x^{J+1}(1-x)^J{2J-1\choose J}-x^{J+1}{2J\choose J}(1-x)^{J+1}\tag{7}
$
Multiply $(6)$ by $x$, multiply $(7)$ by $1-x$, add the results and apply the induction hypothesis to get
$$
x(1-x)(f(x,J)+f(1-x,J))= x(1-x)\tag{8}$$  after verifying that $${2J\choose J}=2{2J-1\choose J}$$
This proves $(2)$ for $x\neq0,1$, but since it is a polynomial identity, it is true for these values also.  
A: Starting from
$$S = \sum_{q=0}^{K-1} {K-1+q\choose K-1}
\frac{a^q b^K + a^K b^q}{(a+b)^{q+K}}$$
we get two pieces
$$\frac{b^K}{(a+b)^K}
\sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^q}{(a+b)^q}
\\ + \frac{a^K}{(a+b)^K}
\sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{b^q}{(a+b)^q}.$$
This is
$$\frac{b^K}{(a+b)^K} 
[z^{K-1}] \frac{1}{1-z} \frac{1}{(1-az/(a+b))^K}
\\ + \frac{a^K}{(a+b)^K} 
[z^{K-1}] \frac{1}{1-z} \frac{1}{(1-bz/(a+b))^K}.$$
Call these $S_1$ and $S_2.$ The first sum is
$$S_1 = \frac{b^K}{(a+b)^K} 
\mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z}
\frac{1}{(1-az/(a+b))^K}
\\ = b^K
\mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z}
\frac{1}{(a+b-az)^K}
\\ = \frac{b^K}{a^K}
\mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z}
\frac{1}{((a+b)/a-z)^K}
\\ = (-1)^{K+1} \frac{b^K}{a^K}
\mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{z-1}
\frac{1}{(z-(a+b)/a)^K}.$$
Now residues sum to zero so we compute this from the residues
at the poles at $z=1$ and $z=(a+b)/a.$ The residue at infinity
is zero by inspection. The residue at $z=1$ is
$$(-1)^{K+1} \frac{b^K}{a^K} \frac{1}{(1-(a+b)/a)^K}
= (-1)^{K+1} b^K  \frac{1}{(a-(a+b))^K}
\\ = (-1)^{K+1}  b^K  \frac{1}{(-b)^K} = -1.$$
For the residue at $z=(a+b)/a$ we require
$$\frac{1}{(K-1)!}
\left(\frac{1}{z^K} \frac{1}{z-1} \right)^{(K-1)}
\\ = \frac{1}{(K-1)!}
\sum_{q=0}^{K-1} {K-1\choose q} 
(-1)^q \frac{(K-1+q)!}{(K-1)!} \frac{1}{z^{K+q}}
(-1)^{K-1-q} \frac{(K-1-q)!}{(z-1)^{K-q}}
\\ = (-1)^{K+1} \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{1}{z^{K+q}}
\frac{1}{(z-1)^{K-q}}.$$
Evaluating the residue we find
$$\left. (-1)^{K+1} \frac{b^K}{a^K}
(-1)^{K+1} \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{1}{z^{K+q}}
\frac{1}{(z-1)^{K-q}} \right|_{z=(a+b)/a}
\\ = \frac{b^K}{a^K}
\sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}}
\frac{1}{((a+b)/a-1)^{K-q}}
\\ = \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}}
\frac{b^q}{a^q}
\frac{b^{K-q}}{a^{K-q}}\frac{1}{((a+b)/a-1)^{K-q}}
\\ = \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}}
\frac{b^q}{a^q}
\\ = \frac{a^K}{(a+b)^K} \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{b^q}{(a+b)^{q}} = S_2.$$
We recognise $S_2$ and hence we have shown that
$$S_1-1+S_2 = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{
\sum_{q=0}^{K-1} {K-1+q\choose K-1}
\frac{a^q b^K + a^K b^q}{(a+b)^{q+K}} = 1}$$
as claimed.
Addendum. This is a special case with $x=a/(a+b)$ of the identity at this MSE link.
