Prove $1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n}} = 2 - \frac{1}{2^{n}}$ for all positive integers $n$.
My approach was to add $\frac{1}{2^{n + 1}}$ to both sides for the induction step. However, I got lost in the algebra and could not figure out the rest of the proof. Any help would be greatly appreciated.
Thank you in advance.
Hint
if we add $\frac{1}{2^{n+1}}$ to the right side, we get
$$2-\frac{1}{2^n}+\frac{1}{2^{n+1}}$$
$$=2-\frac{2}{2^{n+1}}+\frac{1}{2^{n+1}}$$
$$2-\frac{1}{2^{n+1}}$$ qed.
Let $S(n)$ be the statement: $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{n}}=2-\dfrac{1}{2^{n}}$
First do basis step:
$S(1):\hspace{5 mm}$ $1+\dfrac{1}{2^{1}}=\dfrac{3}{2}$
and$\hspace{9 mm}$ $2-\dfrac{1}{2^{1}}=\dfrac{3}{2}$
Then do inductive step:
Assume $S(k)$ is true, or $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{k}}=2-\dfrac{1}{2^{k}}$
$S(k+1):\hspace{5 mm}$ $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{k}}+\dfrac{1}{2^{k+1}}$
$\hspace{18 mm}=2-\dfrac{1}{2^{k}}+\dfrac{1}{2^{k+1}}$
$\hspace{18 mm}=2+\dfrac{1}{2^{k+1}}-\dfrac{1}{2^{k}}$
$\hspace{18 mm}=2+\dfrac{2^{k}-2^{k+1}}{2^{k+1}\cdot{2^{k}}}$
$\hspace{18 mm}=2+\dfrac{2^{k}\big(1-2\big)}{2^{k+1}\cdot{2^{k}}}$
$\hspace{18 mm}=2-\dfrac{1}{2^{k+1}}$
So $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{n}}=2-\dfrac{1}{2^{n}}$.