SOLUTION: prove that a perfact number can be written as a sum of (2^n)-1 consicutive numbers for some n.(please give me the proof)
Algebra.Com
Question 476628: prove that a perfact number can be written as a sum of (2^n)-1 consicutive numbers for some n.(please give me the proof)
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Proving it assumes that all perfect numbers N can be written as
)
, in which this is equal to
(2^n)}{2} = 1 + 2 + 3 + ... + (2^n - 1))
hence we are done. So if mathematicians somehow find an odd perfect number this theorem might not be true.
RELATED QUESTIONS
prove that a perfact number can be written as a sum of (2^n)-1 consicutine numbers for... (answered by richard1234)
Prove that for any integer n, {{{ n(n^2-1)(n+2) }}} is divisible by 12.
Below is my... (answered by ikleyn)
Given that n^2 + n = 2k, how can (n+1)^2 + (n+1) be written as a multiple of... (answered by stanbon)
prove that:
Sum of n numbers in a sequence is... (answered by stanbon,TP)
Please help me solve this induction question using a direct proof:
Prove by induction... (answered by jim_thompson5910)
Can you please explain this problem to me. I have given some of the solutions I have... (answered by jim_thompson5910)
An integer n is called square-free if there does not exist a prime number p such that... (answered by richard1234)
prove that the product of any three consecutive numbers is even.
So far I have tried , (answered by josgarithmetic,tenkun,amalm06,ikleyn)
Decide whether or not the following proposition is true. If it is false, demonstrate this (answered by solver91311,richard1234)