Prove a summation equals to one

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.

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.