Question

Can someone show me how to show that the sum of the divisors of 2^(k-1) is (2^(k+1) -1)

I'm pretty sure you need to use the Proposition for Divisors from Prime Factorization.

I need this for a proof I am doing. Thanks! Can someone show me how to show that the sum of the divisors of 2^(k-1) is (2^(k+1) -1)

I'm pretty sure you need to use the Proposition for Divisors from Prime Factorization.

I need this for a proof I am doing. Thanks! @Mathematics

Answer 1

There aren't very many here who can do these kinds of problems, so look for Zarkon, JamesJ, or Harkirat. I've seen them do this kind of thing befor.

Answer 2

ok

Answer 3

I'm pretty sure the sum of the divisors for 2^(k-1) looks like this
\[\sum_{i=1}^{k - 1} 2i\]

Answer 4

how can I ask them for help?

Answer 5

Look for them on the dashboard. I just checked and the names I mentioned are not available right now, but if you put the cursor over their name and it says "so-and-so is solving some problem" you can click on the problem (in blue) and it will send you to the same page they are looking at. Then you can catch them while they're working and send them the link to your problem.

Answer 6

alright, I don't want to intrude on someone else getting help either though lol

But I will look out for them.

Answer 7

If you post your link just say "can you look at this when you get the chance?" and they'll check it out after they're don with what they're doing.

Answer 8

I just sent Zarkon this problem, hopefully he'll drop in soon.

Answer 9

oh really? thanks!

Answer 10

hmm... he looked at the problem and left. Hope he comes back...

Answer 11

maybe he doesn't know either

Answer 12

it's not correct as stated

Answer 13

I think you need \[2^k-1\]

Answer 14

oh yes you are right

Answer 15

and

\[\sum_{n=0}^{k-1}2^n=2^k-1\]

Answer 16

yes I just noticed that I had put a 1 in there

Answer 17

is there a way that I can break up that summation to show that it equals 2^k -1?

Answer 18

this is a geometric sum..it follows

\[\sum_{n=0}^{k-1}r^n=\frac{r^k-1}{r-1}\]

Answer 19

I suppose there is no way to do it without the geometric sum formula. I haven't been taught that yet so I don't think I can use it.

Answer 20

\[S=1+r+r^2+r^2+\cdots+r^{k-1}\]

\[rS=r+r^2+r^2+\cdots+r^{k}\]

subtract the two

\[S-rS=1-r^k\]

\[S(1-r)=1-r^k\]

\[S=\frac{1-r^k}{1-r}=\frac{r^k-1}{r-1}\]

Answer 21

gotta go...have fun

Answer 22

thanks for your help

Answer 23

Yeah he actually knows what he's talking about. One of the few...

Answer 24

thanks for your help too!