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(D) The function $f(n)$ represents the sum of all positive divisors of $n$. For a number $n = 2^k \cdot 3^1$, the divisors are the products of the div

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The function $f(n)$ represents the sum of all positive divisors of $n$. For a number $n = 2^k \cdot 3^1$, the divisors are the products of the divisors of $2^k$ and the divisors of $3^1$. The sum of divisors of $2^k$ is $(1 + 2 + 2^2 + \dots + 2^k) = \frac{2^{k+1}-1}{2-1} = 2^{k+1}-1$. The sum of divisors of $3^1$ is $(1 + 3) = 4$. Since $f$ is a multiplicative function for coprime factors, $f(2^k \cdot 3) = f(2^k) \cdot f(3)$. 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 $f(n) = \text{the sum of positive divisors of } n$, then $f(2^k \cdot 3)$, where $k$ is a positive integer, is

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