A random variable $X$ takes values $0,1,2,3....$ with probability $P\left( X=x \right)=k\left( x+1 \right){{\left( \dfrac{1}{5} \right)}^{x}}$, where $k$ is a constant, then $P\left( X=0 \right)$ is equal to
$1)\text{ }7/25$
$2)\text{ 18}/25$
$3)\text{ 13}/25$
$4)\text{ 19}/25$
$5)\text{ 16}/25$

Hint: In this question we have been given with a probability function for which the random variable $X$ takes values from $0,1,2,3$ upto infinity. Based on the given probability function we have to find the value of $P\left( X=0 \right)$. We will solve this question by first finding the value of $k$. We will also use the formula of the sum of series which is $a+\left( a+d \right)r+\left( a+d \right){{r}^{2}}+...=\dfrac{a}{1-r}+\dfrac{dr}{{{\left( 1-r \right)}^{2}}}$. We will then substitute $X=0$ and get the required probability.

Complete step-by-step solution:
We have the function given to us as:
$\Rightarrow P\left( X=x \right)=k\left( x+1 \right){{\left( \dfrac{1}{5} \right)}^{x}}$
Now we know that the sum of all the probabilities of an event is $1$ therefore, we can write:
$\Rightarrow \sum\limits_{x=0}^{\infty }{P\left( X=x \right)=1}$
Now from the question we have been given that $P\left( X=x \right)=k\left( x+1 \right){{\left( \dfrac{1}{5} \right)}^{x}}$ therefore, on substituting, we get:
$\Rightarrow \sum\limits_{x=0}^{\infty }{k\left( x+1 \right){{\left( \dfrac{1}{5} \right)}^{x}}=1}$
Now since $k$ is a constant, we can take it out and write the expression as:
$\Rightarrow k\sum\limits_{x=0}^{\infty }{\left( x+1 \right){{\left( \dfrac{1}{5} \right)}^{x}}=1}$
Now on expanding the sum by substituting the values, we get:
$\Rightarrow k\left[ \left( 0+1 \right){{\left( \dfrac{1}{5} \right)}^{0}}+\left( 1+1 \right){{\left( \dfrac{1}{5} \right)}^{1}}+\left( 2+1 \right){{\left( \dfrac{1}{5} \right)}^{2}}+.... \right]=1$
On simplifying, we get:
$\Rightarrow k\left[ 1+\left( 1+1 \right){{\left( \dfrac{1}{5} \right)}^{1}}+\left( 2+1 \right){{\left( \dfrac{1}{5} \right)}^{2}}+.... \right]=1$
Now we know the formula $a+\left( a+d \right)r+\left( a+d \right){{r}^{2}}+...=\dfrac{a}{1-r}+\dfrac{dr}{{{\left( 1-r \right)}^{2}}}$therefore, on substituting $a=1$, $d=1$ and $r=\dfrac{1}{5}$, we get:
$\Rightarrow k\left[ \dfrac{1}{1-\dfrac{1}{5}}+\dfrac{1\times \dfrac{1}{5}}{{{\left( 1-\dfrac{1}{5} \right)}^{2}}} \right]=1$
On taking the lowest common multiple, we get:
\[\Rightarrow k\left[ \dfrac{1-\dfrac{1}{5}+\dfrac{1}{5}}{{{\left( 1-\dfrac{1}{5} \right)}^{2}}} \right]=1\]
on using the expansion of ${{\left( a-b \right)}^{2}}$, we get:
\[\Rightarrow k\left[ \dfrac{1-\dfrac{1}{5}+\dfrac{1}{5}}{1+\dfrac{1}{25}-\dfrac{2}{5}} \right]=1\]
On simplifying, we get:
\[\Rightarrow k\left[ \dfrac{1}{\dfrac{16}{25}} \right]=1\]
On rearranging the terms, we get:
\[\Rightarrow k\left[ \dfrac{25}{16} \right]=1\]
On transferring the terms, we get:
\[\Rightarrow k=\dfrac{16}{25}\]
Therefore, we get the probability function as:
$P\left( X=x \right)=\left( \dfrac{16}{25} \right)\left( x+1 \right){{\left( \dfrac{1}{5} \right)}^{x}}$
Now we have to find $P\left( X=0 \right)$ therefore, on substituting $x=0$, we get:
$P\left( X=0 \right)=\left( \dfrac{16}{25} \right)\left( 0+1 \right){{\left( \dfrac{1}{5} \right)}^{0}}$
On simplifying, we get:
$P\left( X=0 \right)=\dfrac{16}{25}$, which is the required solution.
Therefore, the correct answer is option $\left( 5 \right)$.

Note: It is to be noted that the general principle applied in this question is the sum of all the probabilities of an event is $1$. It is to be remembered that the total probability can never exceed $1$ neither can it be negative. The various series formulas should be remembered to convert an infinite series to a finite sum.