Let $S=\mathbf{N} \cup\{0\}$. Define a relation $R$ from $S$ to $\mathbf{R}$ by:\ $R=\left\{(x, y):

Question

Let $S=\mathbf{N} \cup\{0\}$. Define a relation $R$ from $S$ to $\mathbf{R}$ by:\ $R=\left\{(x, y): \log _{e} y=x \log _{e}\left(\frac{2}{5}ight), x \in S, y \in \mathbf{R}ight\}$.\ Then, the sum of all the elements in the range of $R$ is equal to :\

Professor PC Thomas Classes · Answer

(b): We have, $$ S=N \cup\{0\}=\{0,1,2,3, \ldots \ldots .\} $$ Also, $\log _{e} y=x \log _{e}\left(\frac{2}{5}ight) \Rightarrow y=\left(\frac{2}{5}ight)^{x}$\ $	herefore \quad$ Required sum\ $=\left(\frac{2}{5}ight)^{0}+\left(\frac{2}{5}ight)^{1}+\left(\frac{2}{5}ight)^{2}+\ldots \ldots=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}$.\