Check out some similar questions!

britt_bek_17 · Apr 23, 2009, 06:44 PM

use mathematical induction to prove the statements in these excercises. Note that some of these statements can also be proved be other methods.

n over sigma over I=1, then prove that 2^i-1=(2^n)-1:eek:

Perito · Apr 23, 2009, 08:26 PM

$\frac {n}{(\frac {\sigma}{i})}\, =\, 1$

or

$\frac {(\frac {n}{\sigma}) }{i}\, =\, 1$

I guess what I want to know is what you need to prove. Is it

$\frac {n i}{\sigma} = 1$

or

$\frac {n \sigma}{i} = 1$

------------------------

Then prove that

$(2^i)-1=(2^n)-1$

Is this what you want?

What is σ? Are we talking about the standard deviation here, or something else?

galactus · Apr 24, 2009, 08:06 AM

I believe they mean the capital summation sigma.

$\sum_{i=1}^{n}2^{i-1}=2^{n}-1$

$1+2+2^{2}+2^{3}+.....+2^{n-1}=2^{n}-1$

$P_{1}$ is true because $2^{1-1}=2^{1}-1=1$

Assume true for $P_{k}$

$1+2+2^{2}+2^{3}+.....+2^{k-1}=2^{k}-1$

Show true for $P_{k+1}$

$1+2+2^{2}+2^{3}+.....+2^{k-1}+2^{k}=2^{k}-1+2^{k}$

$=2\cdot 2^{k}-1=2^{k+1}-1$

It is true for the (k+1)st case, therefore, it is true for all cases and it is proven.

galactus · Apr 25, 2009, 06:03 PM

By that problem statement, I can see why you got confused. :confused::rolleyes: