Consider n biased  coins with k^th  coin having probability of throwing head is  $\frac{1}{2\mathrm{k}+1},\mathrm{k}=1,2,3,....,\mathrm{n}.$ All are tossed once together. Let the probability of getting odd number of heads be P_n.  Then answer the following

Let  Pn be probability getting odd number of heads and   be probability of getting even number of heads 
$$\begin{gathered} {\mathrm{P}}_{\mathrm{n}}+{\mathrm{Q}}_{\mathrm{n}}=1 \\ {\mathrm{P}}_{\mathrm{n}}=\frac{1}{2\mathrm{n}+1}{\mathrm{Q}}_{\mathrm{n}-1}+\frac{2\mathrm{n}}{2\mathrm{n}+1}{\mathrm{P}}_{\mathrm{n}-1} \\ {\mathrm{Q}}_{\mathrm{n}}=\frac{1}{2\mathrm{n}+1}{\mathrm{P}}_{\mathrm{n}-1}+{\mathrm{Q}}_{{ }_{\mathrm{n}-1}}\frac{2\mathrm{n}}{2\mathrm{n}+1} \\ {\mathrm{P}}_{\mathrm{n}}-{\mathrm{Q}}_{\mathrm{n}}=\left({\mathrm{P}}_{\mathrm{n}-1}-{\mathrm{Q}}_{{ }_{\mathrm{n}-1}}\right)\frac{2\mathrm{n}-1}{2\mathrm{n}+1} \\ {\mathrm{G}}_{\mathrm{n}}={\mathrm{G}}_{\mathrm{n}-1}\frac{2\mathrm{n}-1}{2\mathrm{n}+1}{\mathrm{G}}_{\mathrm{n}}={\mathrm{G}}_1\frac{3}{2\mathrm{n}+1} \\ {\mathrm{P}}_{\mathrm{n}}-{\mathrm{Q}}_{\mathrm{n}}=\frac{-1}{2\mathrm{n}+1}\&{\mathrm{P}}_{\mathrm{n}}-{\mathrm{Q}}_{\mathrm{n}}=1 \\ {\mathrm{P}}_{\mathrm{n}}=\frac{\mathrm{n}}{2\mathrm{n}+1}{\mathrm{Q}}_{\mathrm{n}}=\frac{2\mathrm{n}}{2\mathrm{n}+1} \end{gathered}$$