I need to prove
$$\sum_{k=0}^n 2^{-k} = \frac{2^{n+1}-1}{2^n}$$
Now I am stuck somewhere in the middle of the proof. I need some help seeing the mistake/trick.
Inductionstart: $$\sum_{k=0}^0 2^{-k} = 1 =\frac{2^{1}-1}{2^0} = 1$$ Inductionrequirement: $$\exists n \in \Bbb{N}:\sum_{k=0}^n 2^{-k} = \frac{2^{n+1}-1}{2^n} $$ Inductionclaim: $$ A(n) \implies A(n+1) $$ $$ =\frac{2^{n+2}-1}{2^{n+1}}$$ Proof: $$\sum_{k=0}^{n+1} 2^{-k} = \sum_{k=0}^{n} 2^{-k} +2^{-k-1} =\frac{2^{n+1}-1}{2^n} + 2^{-n-1}$$ $$=\frac{2^{n+1}-1}{2^n} + \frac{2^{n}*2^{-n-1}}{2^n}$$ $$=\frac{2^{n+1}-1}{2^n} + \frac{2^{-1}}{2^n}$$ $$=\frac{2^{n+1}-1}{2^n} + \frac{2^n}{2}$$ Trying from here to get to the claim $$=\frac{2(2^{n+1}-1)}{2^{n+1}} + \frac{2^{2n}}{2^{n+1}}$$ $$=\frac{2^{n+2}-2 + 2^{2n}}{2^{n+1}}$$
From here I don't see getting to the claim. There must be either something wrong in my steps or in my preparation, but I don't see it.
It happens that $\dfrac{2^{-1}}{2^n}$ is not the same thing as $\dfrac{2^n}2$. Correct that and the rest will follow.