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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
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Verified Answer
The correct answer is:
$3\left(2^{k+1}-1\right)$
Given that $f(x)=$ the sum of positive divisors of $n$.
$\begin{aligned}
\therefore f\left(2^k \cdot 3\right) & =3\left(1+2+2^2+2^3+\ldots+2^k\right) \\
& =3 \frac{\left(2^{-k+1}-1\right)}{2-1}=3\left(2^{-k+1}-1\right)
\end{aligned}$
$\begin{aligned}
\therefore f\left(2^k \cdot 3\right) & =3\left(1+2+2^2+2^3+\ldots+2^k\right) \\
& =3 \frac{\left(2^{-k+1}-1\right)}{2-1}=3\left(2^{-k+1}-1\right)
\end{aligned}$
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