Given relation

The relation is $R=\{(x,y): \ln y = x\ln\left(\frac{2}{5}\right),\ x\in S,\ y\in \mathbb R\}$ where $S=\mathbb N\cup\{0\} = \{0,1,2,3,\dots\}.$

Find the corresponding value of $y$ for each $x$

From $\ln y = x\ln\left(\frac{2}{5}\right),$ exponentiating both sides gives $y = e^{x\ln(2/5)} = \left(\frac{2}{5}\right)^x.$

So the range of $R$ is $\left\{\left(\frac{2}{5}\right)^x : x\in \{0,1,2,3,\dots\}\right\}.$

Thus the elements in the range are $1,\ \frac25,\ \left(\frac25\right)^2,\ \left(\frac25\right)^3,\dots$

Sum of all elements in the range

This is an infinite geometric series with $a=1,\qquad r=\frac25.$

Since $|r|<1$, the sum is $\frac{a}{1-r} = \frac{1}{1-\frac25} = \frac{1}{\frac35} = \frac53.$

Match with options

$\frac53$ corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, they agree.