how come i missed this post
a(n) = 4n+2n+1
we can check that :
a(n)|a(2n) (1)
a(n)|a(4n) (2)
also,
a(kn+3n)-a(kn) = 4kn+3n+2kn+3n-4kn-2kn = 4kn(43n-1) + 2kn(23n-1)
=2kn(23n-1)(...)
also a(n)= (23n-1)/(2n-1)
therefore a(n)|a((k+3)n)-a(kn) (3)
(1)(2) and (3), by induction we...
...we need to show that a(n)|a(kn) when k is not divisible by 3
i know that this is true, maybe rewrite a(kn)/a(n) in a certain way
\frac{(8^{kn}-1)(2^{n}-1)}{(2^{kn}-1)(8^{n}-1)}=\frac{8^{(k-1)n}+...+1}{2^{(k-1)n}+...+1}ok after awhile i don't think this is a good direction
trying induction...
errm
how about (2m-1)|(2n-1) when m|n (also converse )
therefore if n is not divisible by 3, 2n-1 is not divisible by 23-1=7
Also, a(n)= 7*(8n-1+..+1)/(2n-1) is an integer so if 2n-1 is not divisble by 7 then 7|a(n) (when n is not divisible by 3)
When n is not a power of 3 and is divisible by...
:confused: but the right hand side is (8n - 1)/(2n - 1)
i got to that point before, but couldn't do much due to my lack of math skills. Is there some trick here ?
as you have suggested, it appears that 7 divides 1+2^n+4^n when n is not divisible by 3
also 1+2^n+4^n = \frac{7(2^{3(n-1)}+2^{3(n-2)}+...+1)}{ 2^{n -1}+...+1}
still working on it
m(n)=1+2^n+4^n
we need to show that
1. if the prime factorization of n contains any thing different from 3, m(n) would not be prime :
--> somehow we need to factorize:
m((3k+1)p) =
m((3k+2)p) =
where p is the other part in n's prime factorization.
from your suggestion, m(2p) is...
Homework Statement
if 1 + 2^n + 4^n is a prime number then n is a power of 3
Homework Equations
The Attempt at a Solution
If n is not a power of 3, i need to show that we can factorize 1 + 2^n + 4^n
But I am not really sure what should n be if it is not a power of 3 ?
is...
Homework Statement
evaluate
Homework Equations
The Attempt at a Solution
i don't think there is an elementary function as anti derivative for this integral
i tried taylor expansion, doesn't seem to work.
Can anyone give me a hint ?
So you want to deny the existence of buoyancy because it leads to the same reading of total weight ? since we already know that buoyancy exists when body is immersed in water ( in normal case) you should state a solid reason why the buoyancy disappears when the body is at the bottom as i did...
At the bottom of the barrel, if the body is covered in water, then buoyancy affects the apparent weight.
If it is not covered by water ( when the adhesion between the material of the body and water is greater then the adhesion between the water and the barrel) i think there is no buoyancy acts...
Biot Savart's law can be used to calculate magnetic field of a current element or part of a closed loop so i think in the general formula, the integral doesn't have to be on a closed current.