For each $k$, $\mathcal{C}_k$ is biased so that, when tossed, it has probability $\tfrac{1}{(2k+1)}$ of falling heads. If the $n$ coins are tossed, what is the probability that the number of heads is odd? Express the answer as a rational function $n$.
To solve the problem, we need to find the probability that the number of heads is odd when \( n \) biased coins are tossed. Each coin \( \mathcal{C}_k \) has a probability of \(\frac{1}{2k+1}\) of landing heads and \(\frac{2k}{2k+1}\) of landing tails.

1. **Define the probabilities for a single coin:**
   - Probability of heads for coin \( \mathcal{C}_k \): \( p_k = \frac{1}{2k+1} \)
   - Probability of tails for coin \( \mathcal{C}_k \): \( q_k = \frac{2k}{2k+1} \)

2. **Set up the recurrence relations:**
   Let \( O_r \) denote the probability that there are an odd number of heads when \( r \) coins are tossed, and \( E_r \) denote the probability that there are an even number of heads when \( r \) coins are tossed.

   - For \( r = 1 \):
     \[
     O_1 = \frac{1}{3}, \quad E_1 = \frac{2}{3}
     \]

   - For \( r \geq 2 \), the recurrence relations are:
     \[
     O_r = \frac{1}{2r+1}E_{r-1} + \frac{2r}{2r+1}O_{r-1}
     \]
     \[
     E_r = \frac{1}{2r+1}O_{r-1} + \frac{2r}{2r+1}E_{r-1}
     \]

3. **Calculate the probabilities for small values of \( r \):**
   - For \( r = 2 \):
     \[
     O_2 = \frac{1}{5}E_1 + \frac{4}{5}O_1 = \frac{1}{5} \cdot \frac{2}{3} + \frac{4}{5} \cdot \frac{1}{3} = \frac{2}{15} + \frac{4}{15} = \frac{6}{15} = \frac{2}{5}
     \]
     \[
     E_2 = \frac{1}{5}O_1 + \frac{4}{5}E_1 = \frac{1}{5} \cdot \frac{1}{3} + \frac{4}{5} \cdot \frac{2}{3} = \frac{1}{15} + \frac{8}{15} = \frac{9}{15} = \frac{3}{5}
     \]

   - For \( r = 3 \):
     \[
     O_3 = \frac{1}{7}E_2 + \frac{6}{7}O_2 = \frac{1}{7} \cdot \frac{3}{5} + \frac{6}{7} \cdot \frac{2}{5} = \frac{3}{35} + \frac{12}{35} = \frac{15}{35} = \frac{3}{7}
     \]
     \[
     E_3 = \frac{1}{7}O_2 + \frac{6}{7}E_2 = \frac{1}{7} \cdot \frac{2}{5} + \frac{6}{7} \cdot \frac{3}{5} = \frac{2}{35} + \frac{18}{35} = \frac{20}{35} = \frac{4}{7}
     \]

4. **Prove by induction:**
   We hypothesize that:
   \[
   O_n = \frac{n}{2n+1}, \quad E_n = \frac{n+1}{2n+1}
   \]

   **Base case:**
   For \( n = 1 \):
   \[
   O_1 = \frac{1}{3}, \quad E_1 = \frac{2}{3}
   \]
   which matches the hypothesis.

   **Inductive step:**
   Assume the hypothesis holds for \( n = k \):
   \[
   O_k = \frac{k}{2k+1}, \quad E_k = \frac{k+1}{2k+1}
   \]

   For \( n = k+1 \):
   \[
   O_{k+1} = \frac{1}{2(k+1)+1}E_k + \frac{2(k+1)}{2(k+1)+1}O_k
   \]
   \[
   O_{k+1} = \frac{1}{2k+3} \cdot \frac{k+1}{2k+1} + \frac{2k+2}{2k+3} \cdot \frac{k}{2k+1}
   \]
   \[
   O_{k+1} = \frac{k+1}{(2k+3)(2k+1)} + \frac{2k(k+1)}{(2k+3)(2k+1)}
   \]
   \[
   O_{k+1} = \frac{k+1 + 2k(k+1)}{(2k+3)(2k+1)}
   \]
   \[
   O_{k+1} = \frac{k+1(1 + 2k)}{(2k+3)(2k+1)}
   \]
   \[
   O_{k+1} = \frac{(k+1)(2k+1)}{(2k+3)(2k+1)}
   \]
   \[
   O_{k+1} = \frac{k+1}{2k+3}
   \]

   Similarly, for \( E_{k+1} \):
   \[
   E_{k+1} = \frac{1}{2(k+1)+1}O_k + \frac{2(k+1)}{2(k+1)+1}E_k
   \]
   \[
   E_{k+1} = \frac{1}{2k+3} \cdot \frac{k}{2k+1} + \frac{2k+2}{2k+3} \cdot \frac{k+1}{2k+1}
   \]
   \[
   E_{k+1} = \frac{k}{(2k+3)(2k+1)} + \frac{2(k+1)^2}{(2k+3)(2k+1)}
   \]
   \[
   E_{k+1} = \frac{k + 2(k+1)^2}{(2k+3)(2k+1)}
   \]
   \[
   E_{k+1} = \frac{k + 2(k^2 + 2k + 1)}{(2k+3)(2k+1)}
   \]
   \[
   E_{k+1} = \frac{k + 2k^2 + 4k + 2}{(2k+3)(2k+1)}
   \]
   \[
   E_{k+1} = \frac{2k^2 + 5k + 2}{(2k+3)(2k+1)}
   \]
   \[
   E_{k+1} = \frac{(k+1)(2k+2)}{(2k+3)(2k+1)}
   \]
   \[
   E_{k+1} = \frac{k+1}{2k+3}
   \]

   Thus, by induction, the hypothesis holds for all \( n \).

Therefore, the probability that the number of heads is odd when \( n \) coins are tossed is:
\[
\boxed{\frac{n}{2n+1}}
\]