Let . Define a relation R from S to R by: Then, the sum of all the elements in the range of R is equal to: [2025]

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Solution

Q.
Let $S = N \cup {0}$ . Define a relation R from S to R by:

$R = {(x, y) : \log_{e} y = x \log_{e} (\frac{2}{5}), x \in S, y \in R}$

Then, the sum of all the elements in the range of R is equal to: [2025]

1 $\frac{5}{2}$

2 $\frac{5}{3}$

3 $\frac{3}{2}$

4 $\frac{10}{9}$

Ans.
(2)

We have, $S = N \cup {0} = {0, 1, 2, 3, . . . . . .}$

Also, $\log_{e} y = x \log_{e} (\frac{2}{5}) \Rightarrow y = {(\frac{2}{5})}^{x}$

$∴$ Required sum = ${(\frac{2}{5})}^{0} + {(\frac{2}{5})}^{1} + {(\frac{2}{5})}^{2} + . . . . . . . .$

$= \frac{1}{1 - \frac{2}{5}} = \frac{5}{3}$ .

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