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(C) We know that for a probability distribution,the sum of all probabilities must be $1$,i.e.,$\sum_{x=0}^{\infty} P(X=x) = 1$.Substituting the given

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$\frac{16}{25}$

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(C) We know that for a probability distribution,the sum of all probabilities must be $1$,i.e.,$\sum_{x=0}^{\infty} P(X=x) = 1$.Substituting the given expression: $k \sum_{x=0}^{\infty} (x+1) (\frac{1}{5})^x = 1$.Expanding the series: $k [ 1 + 2(\frac{1}{5}) + 3(\frac{1}{5})^2 + 4(\frac{1}{5})^3 + \ldots ] = 1$.This is an arithmetico-geometric series of the form $\sum_{n=0}^{\infty} (a+nd)r^n = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ where $a=1, d=1, r=\frac{1}{5}$.Calculating the sum: $k [ \frac{1}{1 - 1/5} + \frac{1 	imes 1/5}{(1 - 1/5)^2} ] = 1$.$k [ \frac{5}{4} + \frac{1/5}{16/25} ] = k [ \frac{5}{4} + \frac{5}{16} ] = 1$.$k [ \frac{20+5}{16} ] = 1 \Rightarrow \frac{25k}{16} = 1$.Therefore,$k = \frac{16}{25}$.

$X=x$	$-1$	$0$	$1$	$2$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

If a discrete random variable $X$ is defined as follows:
$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$
then $k=$

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If a discrete random variable $X$ is defined as follows:$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$then $k=$