Mathematical Induction:Sum of Reciprocals of Powers of 2

[cyberspace]

Prove for all n is greater than or equal to 1 that:

1/(2^1) + 1/(2^2) + 1/(2^3) + ...+ 1/(2^n) = 1- 1/(2^n)

Here are my steps:
-n=1
1/(2^1) = 1- 1/(2^1)...True
-n=k
1/(2^1) + 1/(2^2) + 1/(2^3) + ...+ 1/(2^k) = 1- 1/(2^k)
-n=k+1
1/(2^1) + 1/(2^2) + 1/(2^3) + ...+ 1/(2^k) + 1/[2^(k+1)] = 1- 1/[2^(k+1)]
1-1/(2^k) + 1/[2^(k+1)] = 1- 1/[2^(k+1)]
And I don't know how to continue from here.

Please notify me if I did a step wrong. Thanks for the help!

 

[galactus]

It would appear you're done. You have it. You showed that

$\displaystyle \frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+....+\frac{1}{2^{k}}+\frac{1}{2^{k+1}}$

$\displaystyle =1-\frac{1}{2^{k}}+\frac{1}{2^{k+1}}=\boxed{1-\frac{1}{2^{k+1}}}$...QED

Good job.