Probability of getting an odd number of heads if n biased coins are tossed once. The question is basically to find out the probability of getting odd number of heads when $ n$ biased coins,with $m^{th}$ coin having probability of throwing head equal to $\frac{1}{2m+1}$ ($m=1,2,\cdots,n$) are tossed once.The results for each coin are independent.
If we consider first that only one head turns up.The probability then is equal to $$\sum_{m=1}^{n} [\frac{1}{2m+1} \prod_{k=1,k \neq m}^{n} (1-\frac{1}{2k+1})]$$ which seems very difficult to evaluate.It gets more complicated if we increase the number of heads.I couldnot find out a simple way to do this.Any suggestion would be highly appreciated.Thanks.

A note for solving lulu's recursion 
We have $$P_{n+1}=P_n \left(1-\frac 2{2n+3}\right)+\frac 1{2n+3}= P_n \frac{2n+1}{2n+3} + \frac 1{2n+3}$$
with $P_0 =0$
Multiplying by $2n+3$:
$$P_{n+1} (2n+3)=P_{n+1}(2 (n+1)+1)=P_n \left({2n+1}\right)+1$$
Calling $A_n = P_n (2n+1)$ this gives
$$A_{n+1}=A_n+1$$
with $A_0=0$. Of course, this implies $A_n=n$ and hence 
$$ P_n = \frac{n}{2n+1}$$
