Let $P_{k}$ be the probability that the total number of heads is odd when $k$ coins are tossed. Therefore, $P_{k+1}$ will be the sum of probability of an odd number of heads until $k$th the coin is tossed multiplied by the probability of getting a tail in the $(k+1)$th toss and the probability of getting an even number of heads until the $k$th the coin is tossed multiplied by the probability of getting a head in the $(k+1)$th toss.
$$
\begin{aligned}
&P_{k+1} = P_{k} \left( \frac{2k+2}{2k+3} \right) + (1-P_{k}) \frac{1}{2k+3} \cr
&(2k+3) P_{k+1} = (2k+1)P_{k} + 1 \cr
&\implies (2k+3) P_{k+1} - (2k+1)P_{k} = 1 \cr
& \implies \sum_{k=0}^{n-1} ( (2k+3) P_{k+1} - (2k+1)P_{k}) = \sum_{k=0}^{n-1} 1 \cr
&\implies (2n+1)P_{n}- 0 = n \ (\text{since} \ P_{0}=0) \cr
&\implies P_{n} = \frac{n}{2n+1}. \cr
\end{aligned}
$$
Method:2 Consider the Polynomial
$$ \prod_{k=1}^{n} \left( \frac{x}{2k+1} + \frac{2k}{2k+1} \right) $$
The sum of coefficients of odd powers of $x$ is equal to the required probability. Say
$$ P(x) = \prod_{k=1}^{n} \left( \frac{x}{2k+1} + \frac{2k}{2k+1} \right) $$
The sum of coefficients of odd powers of $x$ is equal to $\frac{P(1)-P(-1)}{2}.$ Every factor in the product $P(1)$ is $1$, therefore $P(1)$ is $1$.
$$ P(-1) = \prod_{k=1}^{n} \frac{2k-1}{2k+1} = \frac{1}{2n+1}. $$
$$ P_{n} = \frac{1- \frac{1}{2n+1}}{2} = \frac{n}{2n+1}. $$
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