Question

If N denotes set of all positive integers and if f: N→ N is defined by f(n) = the sum of positive divisors of n, then $f\left( {{2}^{k}}.3 \right)$ where ‘k’ is a positive integer is:
(a) ${{2}^{k+1}}-1$
(b) $2\left( {{2}^{k+1}}-1 \right)$
(c) $3\left( {{2}^{k+1}}-1 \right)$
(d) $4\left( {{2}^{k+1}}-1 \right)$

Answer 1

Hint: We know that positive divisors of any positive integer of form ${{a}^{n}}$ is $1,a,{{a}^{2}},{{a}^{3}},......,{{a}^{n}}$. We can see that the obtained divisors are following geometric progression with first term ‘1’ 1 and common ratio ‘r’. We find $f\left( {{2}^{k}}.3 \right)$ by using the distributive property of multiplication $a.b+a.c=a.(b+c)$ between the divisors obtained. Since f(n) is defined as the sum of positive divisors of n, we find the sum of positive divisors of ${{2}^{k}}.3$ by using both distributive property and sum of the geometric progression.

Complete step-by-step solution:
Given that N is a set of all positive Integers and f: N→ N is a function defined as follows:
f(n)=the sum of positive divisors of n.
We need to find the value of $f\left( {{2}^{k}}.3 \right)$, where $k\ge 1$(positive integer).
Let us first find the divisors of ${{2}^{k}}$ and 3 first and find the sum of those divisors later.
We know that the positive divisors of a positive Integer ${{a}^{k}}\left( k\ge 1 \right)$ are $1,a,{{a}^{2}},{{a}^{3}},.......{{a}^{k}}$.
So, the positive divisors of ${{2}^{k}}$ are $1,2,{{2}^{2}},{{2}^{3}},......,{{2}^{k}}$, and the positive divisors of 3 are 1 and 3.
Now the positive divisors of ${{2}^{k}}.3$ will be positive divisors of ${{2}^{k}}$ multiplies by positive divisors of 3.
Therefore the divisors of ${{2}^{k}}.3$ are $(1,2,{{2}^{2}},{{2}^{3}},......,{{2}^{k}}).1$ and \[\left( 1,2,{{2}^{2}},{{2}^{3}},\ldots \ldots {{2}^{k}} \right).3\].
Let us find sum of the divisors $1,2,{{2}^{2}},{{2}^{3}},......,{{2}^{k}}$.
The divisors $1,2,{{2}^{2}},{{2}^{3}},......,{{2}^{k}}$ are in geometric progression.
We know that for a geometric progression $a,ar,a{{r}^{2}},a{{r}^{3}},......,a{{r}^{n}}$ is $\dfrac{a({{r}^{n+1}}-1)}{r-1}$, where ‘a’ is first term and ‘r’ is common ratio.
Let us assume sum of $1,2,{{2}^{2}},{{2}^{3}},......,{{2}^{k}}$ be $Su{{m}_{2}}$. Here the first term is ‘1’ and the common ratio is 2.
$Su{{m}_{2}}=1+2+{{2}^{2}}+{{2}^{3}}+......+{{2}^{k}}$
$Su{{m}_{2}}=\dfrac{{{2}^{k+1}}-1}{2-1}$
$Su{{m}_{2}}=\dfrac{{{2}^{k+1}}-1}{1}$
$Su{{m}_{2}}={{2}^{k+1}}-1.......(1)$
We know that $a.b+a.c=a.(b+c)$.
From equation (1) we got the sum of divisors of ${{2}^{k}}$. We multiply $Su{{m}_{2}}$ with each divisor of 3.
Sum of divisors of ${{2}^{k}}.3$ is $\left( {{2}^{k+1}}-1 \right).1+\left( {{2}^{k+1}}-1 \right).3$
Sum of divisors of ${{2}^{k}}.3$ is $\left( {{2}^{k+1}}-1 \right).(1+3)$
Sum of divisors of ${{2}^{k}}.3$ is $({{2}^{k+1}}-1).4$.
$\therefore$ The value of $f\left( {{2}^{k}}.3 \right)$ is $({{2}^{k+1}}-1).4$. The correct option is D.

Note: Mistakes may arise while calculating the sum for the positive divisors of ${{2}^{k}}$ as we have k+1 positive divisors here. We know that the sum of ‘n’ terms of geometric progression is \[\dfrac{a\left( {{r}^{n-1}}-1 \right)}{r-1}\] as we have ‘k+1’ we need to replace ‘n’ with ‘k+1’ not with k. Since ‘1’ is a common divisor for both ${{2}^{k}}$ and 3 it ensures that all positive divisors are included.