If a discrete random variable \(X\) is defined as follows
\(\mathrm{P}[\mathrm{X}=x]=\left\{\begin{array}{cl}
\frac{\mathrm{k}(x+1)}{5^x}, & \text { if } x=0,1,2 \ldots \ldots \
0, &
\end{array}\right.\) \(\text { otherwise then }\)\(\mathrm{k}=\)

We have, \(\sum_{x=0}^{\infty} \mathrm{P}(\mathrm{X}=x)=1\)
\(\begin{aligned}
& \Rightarrow \mathrm{k} \sum_{x=0}^{\infty}(x+1)\left(\frac{1}{5}\right)^x=1 \\
& \Rightarrow \mathrm{k}\left[1+2\left(\frac{1}{5}\right)+3\left(\frac{1}{5}\right)^2+4\left(\frac{1}{5}\right)^3+\ldots\right]=1
\end{aligned}\)
\(\Rightarrow \mathrm{k}\left[\frac{1}{1-\frac{1}{5}}+\frac{1 \times \frac{1}{5}}{\left(1-\frac{1}{5}\right)^2}\right]=1\)
\(\cdots\left[\begin{array}{r}\because a+(a+d) r+(a+2 d) r^2+\ldots . \\ =\frac{a}{1-r}+\frac{d r}{(1-r)^2}\end{array}\right]\)
\(\begin{aligned} & \Rightarrow \mathrm{k}\left(\frac{5}{4}+\frac{5}{16}\right)=1 \\ & \Rightarrow \frac{25 \mathrm{k}}{16}=1 \\ & \Rightarrow \mathrm{k}=\frac{16}{25}\end{aligned}\)