Proving 2^n - 1 Formula for Positive Integers | Induction Question

[zeion]

Homework Statement

Show that the statement holds for all positive integers n

2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{n-1} = 2^n - 1

Homework Equations

The Attempt at a Solution

Assume:
2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1} = 2^k - 1 is true.

Then:
2^{k+1}-1 = 2^k(2) - 1 = 2(2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1})

Not sure what to do next :/

Show that:
2^k(2) - 1 = 2(2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1}) is true

 

[tiny-tim]

Hi zeion!

You have the right basic idea, but you're making it too complicated.

Do the obvious …

just multiply both sides of the original equation by 2 (and then fiddle about with it).

 

[zeion]

Haven't I already proved it?

Since:
<br /> 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1} = 2^k - 1 <br />

Then:
<br /> (2)(2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1}) = (2)2^k - 1 = 2^{k+1}-1<br />

 

[tiny-tim]

No, it isn't … be careful! …

(2)(2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1}) = (2)(2^k - 1) = \cdots\ ?