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:

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

or

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

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

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

or

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

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

Then prove that

\Large (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?

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.

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