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 $\sum_{i=1}^n a_i$.
compute the fraction of products
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combinatorics
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0@Ross Millikan: But suppose that $\frac{A_i}{B_i}=1-\frac{1}{B_i}$ as you claim, then $A_i=B_i-1$. Take $n=3$. $A_3=B_3-1$ $\Longrightarrow$ $K_1 \cdot K_2=(K_1+1)\cdot(K_2+1)-1$ $\Longrightarrow$ $K_1K_2=K_1K_2+K_1+K_2+1-1$ $\Longrightarrow$ $K_1+K_2=0$ $\Longrightarrow$ $2\cdot 2^2+2\cdot 2^1=2$ which is false. – 2011-03-19
1 Answers
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$\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).