Sums

Expand the series for each value of $n$ : $\frac {1}{2^0}$ + $\frac {1}{2^1}$ + $\frac {1}{2^2}$ + $\frac {1}{2^3}$ + $\frac {1}{2^4}$ + $\frac {1}{2^5}$ + $\frac {1}{2^6}$ + $\frac {1}{2^7}$ + $\frac {1}{2^8}$

Simplify each term: $\frac {1}{1}$ + $\frac {1}{2}$ + $\frac {1}{4}$ + $\frac {1}{8}$ + $\frac {1}{16}$ + $\frac {1}{32}$ + $\frac {1}{64}$ + $\frac {1}{128}$ + $\frac {1}{256}$

Find the common denominator: $\frac {256}{256}$ + $\frac {1 \cdot 128}{256}$ + $\frac {1 \cdot 64}{256}$ + $\frac {1 \cdot 32}{256}$ + $\frac {1 \cdot 16}{256}$ + $\frac {1 \cdot 8}{256}$ + $\frac {1 \cdot 4}{256}$ + $\frac {1 \cdot 2}{256}$ + $\frac {1}{256}$

Combine the fractions: $\frac {256+1\cdot 128+1\cdot 64+1\cdot 32+1\cdot 16+1\cdot 8+1\cdot 4+1\cdot 2+1}{256}$

Simplify the numerator: $\frac {511}{256}$

Simplify again: 1 $\frac {255}{256}$

$\displaystyle \sum_{n=0}^8 \dfrac{1}{2^n} = 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \ldots + \dfrac{1}{2^8}$

This is a sum of a geometric progression where $a=1,\;r=\dfrac{1}{2}$ and $n=9$

Therefore,

$S_9 = \dfrac{1\left(1-\left(\frac{1}{2}\right)^9\right)}{1-\frac{1}{2}}\\ =\dfrac{1\left(1-\frac{1}{512}\right)}{\frac{1}{2}}\\ =2\left(\dfrac{511}{512}\right)\\ =\dfrac{1022}{512}\\ =\dfrac{511}{256}\\ =\boxed{1\dfrac{255}{256}}$