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(D) The function $f(n)$ represents the sum of all positive divisors of $n$. For a number $n = p_1^{a_1} \cdot p_2^{a_2} \cdots$,the sum of divisors is

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$4(2^{k+1}-1)$

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(D) The function $f(n)$ represents the sum of all positive divisors of $n$. For a number $n = p_1^{a_1} \cdot p_2^{a_2} \cdots$,the sum of divisors is given by $\sigma(n) = \frac{p_1^{a_1+1}-1}{p_1-1} \cdot \frac{p_2^{a_2+1}-1}{p_2-1} \cdots$. Given $n = 2^k \cdot 3^1$,the sum of divisors is: $f(2^k \cdot 3) = (1 + 2 + 2^2 + \cdots + 2^k) \cdot (1 + 3)$. Using the geometric series sum formula $\sum_{i=0}^{k} 2^i = \frac{2^{k+1}-1}{2-1} = 2^{k+1}-1$. Thus,$f(2^k \cdot 3) = (2^{k+1}-1) \cdot (4) = 4(2^{k+1}-1)$.

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 $f(2^k \cdot 3)$,where $k$ is a positive integer,is

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