compute the fraction of products Let $A_i=K_1 \times K_2 \times K_3 \times \cdots \times K_{i-1}$ and $B_i=(K_1+1) \times (K_2+1) \times (K_3+1) \times \cdots \times (K_{i-1}+1)$, where $K_{i-1}=(n-1)2^{n-(i-1)}-1$, $i=2,3,\cdots$. Find $a_i=\frac{A_i}{B_i}$ and $\sum_{i=1}^n a_i$.

$$\begin{align}
a_i&=\frac{A_i}{B_i}=\prod_{m=1}^{i-1}\left(1-\frac{1}{K_m+1}\right)\\
&=\prod_{m=1}^{i-1}\left(1-\frac{1}{(n-1)2^{n-m}}\right)\\
&=\prod_{m=1}^{i-1}\left(1-\frac{1}{(n-1)2^n}2^m\right)\\
&=\left(\frac{1}{(n-1)2^{n-1}},2\right)_{i-1},
\end{align}
$$
where I use the notation for the q-Pochhammer symbol.
So what you want is the sum $\sum_{i=0}^{n-1}(c_n,q)_{i}$ where $c_n=\frac{1}{(n-1)2^{n-1}}$.
This is exactly a situation for the $q$-Zeilberger algorithm to find a recurrence for the sum. If this works you can use the $q$-Hyper algorithm to check if there is a $q$-hypergeometric solution (which is not very likely because the sum does have two unnatural bounds, so it will be a ${}_3\Phi_2$ that is not Pfaff-Saalschütz).
