Let $o(n)$ be the number of $2n-$tuples $(a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n)$ such that each $a_i,b_j\in\{0,1\}$ and $a_1b_1+a_2b_2+\ldots +a_nb_n$ is odd. Similarly, let $e(n)$ be the number of $2n-$tuples such that the sum is even. Find $\frac{o(n)}{e(n)}$ in terms of $n.$
$\textbf{My Approach:}$
If a tuple has an odd sum, let it be called an "odd tuple". Conversely, if it has an even sum, let it be called an "even tuple". Let $(a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n)$ be an odd tuple. Then, $(a_{n+1},b_{n+1})=(0,0),(0,1),(1,0),(1,1).$ But, the tuples $(0,0),(0,1),(1,0)$ when "appended" onto the original tuple don't change the parity of the given sum. Hence, each odd tuple gives rise to three more odd tuples. If the original tuple is even, the appending of $(1,1)$ changes the parity of the sum. Hence, each even tuple gives rise to one odd tuple. So, we get that $o(t+1)=3o(t)+e(t)\forall t\in\mathbb{N}.$ Using very similar reasoning, $e(t+1)=3e(t)+o(t)\forall t\in\mathbb{N}.$ We may compute $o(1)$ and $e(1)$ manually. We have that $o(1)=1$ and $e(1)=3.$ Now, my claim is that $r(n)=\frac{o(n)}{e(n)}=\frac{2^n-1}{2^n+1}\forall n\in\mathbb{N}.$ I proceed by induction. The base case holds true as $r(1)=\frac{1}{3}.$ Let it be true for some $t\in\mathbb{N}.$ Then, $r(t+1)=\frac{o(t+1)}{e(t+1)}=\frac{3o(t)+e(t)}{3e(t)+o(t)}.$ We divide throughout by $e(t)$ to get $r(t+1)=\frac{3r(t)+1}{3+r(t)}=\frac{2^{t+1}-1}{2^{t+1}+1}.$ By induction, $r(n)=\frac{2^n-1}{2^n+1}\forall n\in\mathbb{N}.$
Is my method right? I am unsure as this was on a test, and someone else who used the same method as I did and arrived at the same answer as I did got $2/10$ for this question. Why could this be? The only step that may be problematic is the "appending" step, in my opinion.