Question

If \[n = {2^{p - 1}}\left( {{2^p} - 1} \right)\] , where \[\left( {{2^p} - 1} \right)\] is a prime. Then what is the sum of the divisors of \[n\]?
A. \[n\]
B. \[2n\]
C. \[pn\]
D. \[{p^n}\]

Answer 1

Hint: Here, a number is given. First, using the terms of the number find each divisor. After that, calculate the sum of the divisors. Simplify the equation of the sum of the divisors by using the formula of the sum of the finite terms in the geometric progression. In the end, solve the equation by using the power properties of a number and get the required answer.

Formula Used: \[{a^m}{a^n} = {a^{m + n}}\]
The sum of the first \[n\] terms in the geometric progression \[a,ar,a{r^2},...,a{r^{n - 1}}\] is: \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\] if \[r \ne 1\] . Where \[a\] is the first term and \[r\] is the common ratio.

Complete step by step solution: Given:
\[n = {2^{p - 1}}\left( {{2^p} - 1} \right)\] and \[\left( {{2^p} - 1} \right)\] is a prime number.
Let’s find out the divisors of the given number \[n\].

Let consider \[n = {2^{p - 1}}q\], where \[{2^p} - 1 = q\].
So, the divisors of \[n = {2^{p - 1}}q\] are
\[1,2,{2^2},{2^3},....,{2^{p - 1}},q,2q,{2^2}q,{2^3}q,...{2^{p - 2}}q,{2^{p - 1}}q\]
The sum of the divisors of \[n\] is:
\[S = \left( {1 + 2 + {2^2} + {2^3} + .... + {2^{p - 1}} + q + 2q + {2^2}q + {2^3}q + ... + {2^{p - 2}}q + {2^{p - 1}}q} \right)\]
Solve the right-hand side of the above equation.
\[S = \left( {1 + 2 + {2^2} + {2^3} + .... + {2^{p - 1}}} \right) + \left( {q + 2q + {2^2}q + {2^3}q + ... + {2^{p - 2}}q + {2^{p - 1}}q} \right)\]
\[ \Rightarrow S = \left( {1 + 2 + {2^2} + {2^3} + .... + {2^{p - 1}}} \right) + q\left( {1 + 2 + {2^2} + {2^3} + ... + {2^{p - 2}} + {2^{p - 1}}} \right)\]
The terms of the above summation are in the geometric progression with the common difference 2.
So, apply the formula of the sum of the terms in a geometric progression on the right-hand side.
We get,
\[S = \dfrac{{1\left( {{2^p} - 1} \right)}}{{2 - 1}} + q\dfrac{{1\left( {{2^p} - 1} \right)}}{{2 - 1}}\]
\[ \Rightarrow S = \left( {{2^p} - 1} \right) + q\left( {{2^p} - 1} \right)\]
\[ \Rightarrow S = \left( {{2^p} - 1} \right)\left( {1 + q} \right)\]
Substitute the value of \[q\] in the above equation.
\[S = \left( {{2^p} - 1} \right)\left( {1 + {2^p} - 1} \right)\]
\[ \Rightarrow S = \left( {{2^p}} \right)\left( {{2^p} - 1} \right)\]
\[ \Rightarrow S = 2\left( {{2^{p - 1}}} \right)\left( {{2^p} - 1} \right)\]
\[ \Rightarrow S = 2n\]
Therefore, the sum of the divisors of \[n = {2^{p - 1}}\left( {{2^p} - 1} \right)\] is \[2n\].

Option ‘B’ is correct

Note: Students get confused about the formulas of the sum of the terms in the geometric progression:
The sum of the first \[n\] terms in the geometric progression \[a,ar,a{r^2},...,a{r^{n - 1}}\] is
Formulas:
\[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\], if \[r \ne 1\] and \[r > 1\]
\[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\], if \[r \ne 1\] and \[r < 1\]