I need to prove by mathematical induction:
$$ \frac{1}{2} + \frac{1}{2^2} + ... + \frac{1}{2^n} = 1 - \frac{1}{2^n} .$$
My base case is: Assume $n = 1$
$$ \frac{1}{2} = 1 - \frac{1}{2^1} = 1 - \frac{1}{2} = \frac{1}{2} .$$
My inductive hypothesis is: Assume $n = k$
$$ \frac{1}{2} + \frac{1}{2^2} + ... + \frac{1}{2^k} = 1 - \frac{1}{2^k} .$$
we show K+1 is true
$$ \frac{1}{2} + \frac{1}{2^2} +... + \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
we know
$$ \frac{1}{2} + \frac{1}{2^2} +... + \frac{1}{2^k} = 1 - \frac{1}{2^k} .$$
so
$$ 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
then i go on to manipulate the expression by
$$ 1 -\frac{1*2}{2^{k}*2} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
$$ 1 \frac{-2}{2^{k+1}} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
$$ 1 \frac{-2+1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
$$ 1 - \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
Update on final stage. Would this make more sense? I am getting very confused on how to manipulate these fractions.
$$ 1 -\frac{1*2}{2^{k}*2} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
$$ 1 -\frac{2}{2^{k+1}} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
$$ 1 - \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} .$$
but i dont think last bit is right. I'm not sure if i should swau the minus sign to the numerator when trying to multiply by 2 to make it so there is a common denominator I dont think i am manipulating the exponents in the denominator correctly. Does anybody have any ideas?
Induction: Assume $\sum\limits_{k=1}^n\frac{1}{2^k}=1-\frac{1}{2^n}$ then $\sum\limits_{k=1}^{n+1}\frac{1}{2^k}=1-\frac{1}{2^n}+\frac{1}{2^{n+1}}=1-\frac{1}{2^{n+1}}$.