Question

If $N$ denotes set of all positive integers and if $f:N⟶N$ is defined by $f\left(n\right)$ $=$ the sum of positive divisors of $n$, then $f\left(2^k.3\right)$ where ${}^{\prime }{k}^{\prime }$ is a positive integer is

• A. $2^{k+1}-1$
• B. $2(2^{k+1}-1)$
• C. $3(2^{k+1}-1)$
• D. $4(2^{k+1}-1)$

Answer 1

The correct option is D $4(2^{k+1}-1)$

The divisors of $\left(2^k\right)\left(3\right)$ are $1,2,2^2,.......,2^k,3,3\times 2,3\times 2^2,.......,3\times 2^k$

If you think about it, the sum of all thepositiv divisors of ${(2}^k\left)\right(3)$ can be conveniently written as:
$(2^0+2^1+2^2+...+2^k)(3^0+3^1)$

Now, $2^0,2^1,2^2,........,2^k$ forms a geometric series.

Where $a=2^0=1,r=2$
Using the formula for the sum of a GP:
$S_n=\frac{a(r^n-1)}{(r-1)}$
$=\frac{1(2^{k+1}-1)}{(2-1)}\times (1+3)$

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