Question 475361
First, we have to assume that all perfect numbers N can be written as *[tex \LARGE N = (2^{k-1})(2^k - 1)] for positive integer k.


Let m, m+1, ..., (m+2^n)-2 be the (2^n) - 1 consecutive numbers, in which their sum is


*[tex \LARGE \frac{(2^n - 1)(2m + 2^n - 2)}{2} = (2^n - 1)(m + 2^{n-1} - 1)]


We want to show that all perfect numbers N can be expressed in this form. However, if we set m = 1 the solution becomes trivial.


Alternatively, we can let *[tex \LARGE N = \frac{(2^k - 1)(2^k)}{2}] in which N becomes the sum of the first (2^k) - 1 integers.