If $N$ denotes the set of all positive integers and if $f: N \rightarrow N$ is defined by $f(n)=$ the sum of…
If $N$ denotes the set of all positive integers and if $f: N \rightarrow N$ is defined by $f(n)=$ the sum of positive divisors of $n$ then, $f\left(2^k \cdot 3\right)$, where $k$ is a positive integers, is
$2^{k+1}-1$
$2\left(2^{k+1}-1\right)$
$3\left(2^{k+1}-1\right)$
$4\left(2^{k+1}-1\right)$
Solution
Given that $f(x)=$ the sum of positive divisors of $n$.
$\begin{aligned}
\therefore f\left(2^k \cdot 3\right) & =3\left(1+2+2^2+2^3+\ldots+2^k\right) \\
& =3 \frac{\left(2^{-k+1}-1\right)}{2-1}=3\left(2^{-k+1}-1\right)
\end{aligned}$