If the range of a random variable X is {0,1,2, 3, 4......} with P(x=k) =$\frac{(\mathrm{k}+1)\mathrm{a}}{3^{\mathrm{k}}}\text{ for }\mathrm{k}\ge 0$then a =

$$\begin{gathered} \sum _{\mathrm{k}=0}^{\mathrm{\infty }} \mathrm{P}(\mathrm{x}=\mathrm{k})=1\Rightarrow \sum _{\mathrm{k}=0}^{\mathrm{\infty }} \frac{(\mathrm{k}+1)\mathrm{a}}{3^{\mathrm{k}}}=1 \\ \Rightarrow a+\frac{2a}{3}+\frac{3a}{3^2}+\frac{4a}{3^3}+.....=1 \\ \Rightarrow 1+2\left(\frac{1}{3}\right)+3\left(\frac{1}{3^2}\right)+4\left(\frac{1}{3^3}\right)+....=\frac{1}{a} \\ \Rightarrow \frac{1}{1-{\displaystyle \frac{1}{3}}}+\frac{1\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}{{\left(1-\frac{1}{3}\right)}^2}=\frac{1}{a}\Rightarrow \frac{3}{2}+\frac{3}{4}=\frac{1}{a} \\ \left[In A.G.P, sum of infinite terms=\frac{a}{1-r}+\frac{dr}{{\left(1-r\right)}^2}\right] \\ \Rightarrow a=\frac{4}{9} \end{gathered}$$