(a) Show that $𝜎 (2^n) = 2^{n+1} − 1$.
(b) Show that $𝜎 (p^n) = p^{n+1} − 1$ where $p$ is prime.
(a) The sigma function of a natural number is the sum of all the positive divisors of that number.
The divisors of $2^n$ are $1, 2, 2^2, 2^3, \ldots , 2^n$. And so
$$ \begin{align} \sigma(2^n) & = 1 + 2 + 2^2 + 2^3 \ldots 2^n \\ \\ & = \frac{1-2^{n+1}}{1-2} \quad \text{geometric sum} \\ \\ & = 2^{n+1} - 1 \end{align}$$
Note: the formula for the sum of a geometric series is here (link).
(b) The assertion is wrong. Consider $p=3$ as a counter-example.
The divisors of $3^n$ are $1, 3, 3^2, 3^3, \ldots , 3^n$. And so
$$ \begin{align} \sigma(3^n) & = 1 + 3 + 3^2 + 3^3 \ldots 3^n \\ \\ & = \frac{1-3^{n+1}}{1-3} \quad \text{geometric sum} \\ \\ & = \frac{3^{n+1} - 1}{2} \\ \\ & \ne 3^{n+1}-1 \end{align}$$
The correct expression is
$$ \sigma(p^n) = \frac{p^{n+1}-1}{p-1}$$