Question

Let \( S = \mathbb{N} \cup \{0\} \). Define a relation \( R \) from \( S \) to \( \mathbb{R} \) by: \[ R = \left\{ (x, y) : \log_e y = x \log_e \left(\frac{2}{5}\right), x \in S, y \in \mathbb{R} \right\}. \] Then, the sum of all the elements in the range of \( R \) is equal to:

• A. \( \frac{3}{2} \)
• B. \( \frac{5}{3} \)
• C. \( \frac{10}{9} \)
• D. \( \frac{5}{2} \)

Hint:

In geometric series problems, ensure to correctly identify the first term \( a \) and the common ratio \( r \). The infinite series converges only if \( |r|<1 \).

Answer 1

The Correct Option is B

Step 1: Identifying the relation. From the given relation, we have: \[ \log_e y = x \log_e \left(\frac{2}{5}\right) \] Exponentiating both sides: \[ y = \left(\frac{2}{5}\right)^x \]

Step 2: Summing the range values. The range values are given by the infinite series: \[ \text{Sum} = 1 + \left(\frac{2}{5}\right) + \left(\frac{2}{5}\right)^2 + \left(\frac{2}{5}\right)^3 + \cdots \] This is a geometric progression with first term \(a = 1\) and common ratio \(r = \frac{2}{5}\).

Step 3: Summing the infinite series. Using the sum formula for an infinite geometric progression: \[ \text{Sum} = \frac{a}{1 - r} = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]