Show that \[n={{2}^{m-1}}\left( {{2}^{m}}-1 \right)\] is a perfect number, if \[\left( {{2}^{m}}-1 \right)\] is a prime number.

Hint: Now we are given that \[\left( {{2}^{m}}-1 \right)\] is a prime number. Hence we will write the number as p.
Now substituting p in the equation for n we will list all the factors of n. Now we will take the sum of all the factors and use the sum of GP to evaluate the sum. Note that the sum of GP is given by $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ . Now we will simplify and substitute the value of p and hence easily prove the sum to be equal to n.

Complete step-by-step solution:
Let us first understand the concept of perfect numbers.
Perfect numbers are numbers whose sum of all the proper divisors is equal to the number itself.
For example number 6. The proper divisors of 6 are 1, 2, and 3. Now the sum of its proper divisors are 1 + 2 + 3 = 6. Hence we have the sum of its proper divisors is equal to the number itself.
Hence we can say that 6 is a perfect number.
Now let us consider the given number.
\[n={{2}^{m-1}}\left( {{2}^{m}}-1 \right)\]
We know that ${{2}^{m}}-1$ is a prime number so let us say $p={{2}^{m}}-1$ .
Hence we have $n={{2}^{m-1}}\times p$
Now let us note all the proper factors of n.
$1\times 2,{{2}^{2}},{{2}^{3}},{{2}^{4}},{{...........2}^{m-1}},p,2p,{{2}^{2}}p,{{2}^{3}}p,{{......2}^{m-2}}p$ .
Now let us take the sum of all the factors.
$\begin{align}
&S=1+2+{{2}^{2}}+{{2}^{3}}+{{2}^{4}}+..........+{{2}^{m-1}}+p+2p+{{2}^{2}}p+{{2}^{3}}p+.....+{{2}^{m-2}}p \\
 & S=\left( 1+2+{{2}^{2}}+{{2}^{3}}+{{2}^{4}}+..........+{{2}^{m-1}}+p\left( 1+2+{{2}^{2}}+{{2}^{3}}+{{2}^{4}}+..........+{{2}^{m-2}} \right) \right).......................\left( 1 \right) \\
\end{align}$
Now first consider the series $1+2+{{2}^{2}}+..........+{{2}^{m-1}}$
The series is a GP of m terms with a = 1, r = 2.
Now we know that the sum of GP is given by $\dfrac{a\left( {{r}^{m}}-1 \right)}{r-1}$ .
Hence we get, $1+2+{{2}^{2}}+{{.......2}^{m-1}}=\dfrac{1\left( {{2}^{m}}-1 \right)}{2-1}$
$1+2+{{2}^{2}}+{{.......2}^{m-1}}={{2}^{m}}-1........................\left( 2 \right)$
And similarly we have
 \[1+2+{{2}^{2}}+{{.......2}^{m-2}}=\dfrac{1\left( {{2}^{m-1}}-1 \right)}{2-1}\]
\[\begin{align}
  & p\left( 1+2+{{2}^{2}}+{{.......2}^{m-2}} \right)=\dfrac{p\left( {{2}^{m-1}}-1 \right)}{2-1} \\
& p\left( 1+2+{{2}^{2}}+{{.......2}^{m-2}} \right)=p\left( {{2}^{m-1}}-1 \right).........................\left( 3 \right) \\
\end{align}\]
Now from equation (1), equation (2) and equation (3) we get,
$\begin{align}
  & S={{2}^{m}}-1+p\left( {{2}^{m-1}}-1 \right) \\
 & \Rightarrow S={{2}^{m}}-1+p{{2}^{m-1}}-p \\
 & \Rightarrow S={{2}^{m-1}}\left( 2+p \right)-1\left( 1+p \right) \\
\end{align}$
Now resubstituting the value of p which is $p={{2}^{m}}-1$ we get,
$\begin{align}
  & S={{2}^{m-1}}\left( 2+{{2}^{m}}-1 \right)-1\left( 1+{{2}^{m}}-1 \right) \\
 & \Rightarrow S={{2}^{m-1}}\left( {{2}^{m}}+1 \right)-{{2}^{m}} \\
 & \Rightarrow S={{2}^{m-1}}\left( {{2}^{m}}+1-2 \right) \\
 & \Rightarrow S={{2}^{m-1}}\left( {{2}^{m}}-1 \right) \\
\end{align}$
But we have \[n={{2}^{m-1}}\left( {{2}^{m}}-1 \right)\] hence we get
$S=n$
Hence we have the sum of all the proper divisors of n is equal to n. Hence n is a perfect number.

Note: Now note that for showing a number as a perfect number we take the sum of all the proper divisors of n, which means we should exclude the number itself. Hence when we listed out all the divisors of n we did not include the number ${{2}^{m-1}}p$.