Is this proof done wrongly? I have tried for hours at this problem, but feel as if there is no way to prove it. Isn't this just false?
n > 0.
$$\sum_{k=0}^n \frac{1}{2^k} = 2 = \frac{1}{5^n}$$
I keep getting false because 2(left hand sided answer) is NOT equal to 3/2 (which is what I am getting on the right hand side) with the assumption that n =1. Am I attempting this problem correctly?
 A: You're not starting out the problem correctly. Whenever you do a proof by induction, your base case should always be the lowest number in the set you're working with. In this case, we're working with $n \geq 0$. The lowest number in that set of nonnegative integers is $n=0$, so that's what your base case should be about. (Surely you can start at $n=1$, but then you'd also have to make a case where $n=0$ as well.)
Also, $n=1$ works because
$$\displaystyle\sum_{k=0}^{1} \frac{1}{2^k} = \frac{1}{2^0} + \frac{1}{2^1} = \frac{3}{2}$$
and
$$2 - \frac{1}{2^1} = \frac{3}{2}.$$
(Answer) The given statement you're trying to prove is true.
Base Step. If $n=0$, then
$$\sum_{k=0}^{0}\frac{1}{2^{k}}=\frac{1}{2^{0}}=1$$
and
$$2-\frac{1}{2^{0}} = 1.$$
Inductive Step. Fix some nonnegative integer $m \geq 0$ and suppose
$$\sum_{k=0}^{m}\frac{1}{2^{k}}=2-\frac{1}{2^{m}}.$$
Then
$$\sum_{k=0}^{m+1}\frac{1}{2^{k}}=\sum_{k=0}^{m}\frac{1}{2^{k}}+\frac{1}{2^{m+1}}\ =\ 2-\frac{1}{2^{m}}+\frac{1}{2^{m+1}}.$$
I think you can take it from here. Does that help?
A: The first few sums look like this:
$$\sum_{k=0}^0 \frac{1}{2^k} = 1=2-{1\over 2^0}$$
$$\sum_{k=0}^1 \frac{1}{2^k} = 1+\frac12=\frac32=2-{1\over 2^1}$$
$$\sum_{k=0}^2 \frac{1}{2^k} = 1+\frac12+\frac14=\frac74=2-{1\over 2^2}$$
$$\sum_{k=0}^3 \frac{1}{2^k} = 1+\frac12+\frac14+\frac18=\frac{15}8=2-{1\over 2^3}$$
So the thing you're being asked to prove seems to be correct.
Can you fill in the induction step yourself? It is quite simple.
A: Let's apply the binary system. We want to add   $$1.\underbrace{11\ldots 11}_{n \ {\rm digits}}+0.\underbrace{00\ldots 01}_{n \ {\rm digits}}$$
We obtain
$${\displaystyle\phantom{+}1.11\ldots 11\atop + \ \displaystyle\underline{0.00\ldots 01}}\atop \hspace{3pt}10.00\ldots 00$$
Hence $$\sum_{i=0}^n{1\over 2^i}+{1\over 2^n}=2$$
