Preview Activity 8.3.1.
Let’s explore sums of the form
\begin{equation*}
S_n = 1 + \frac{1}{2} + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^3 + \cdots + \left( \frac{1}{2} \right)^{n-1}\text{.}
\end{equation*}
Observe that we can think of the first term, \(1\text{,}\) as \((\frac{1}{2})^0\text{,}\) and since the powers of \(\frac{1}{2}\) include every whole number from \(0\) to \(n-1\text{,}\) there are exactly \(n\) terms in the sum given by \(S_n\text{.}\)
(a)
Note that \(S_1 = 1\text{,}\) \(S_2 = S_1 + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}\text{,}\) and \(S_3 = S_2 + \frac{1}{4} = \frac{3}{2} + \frac{1}{4} = \frac{7}{4}\text{.}\) Using the fact that each subsequent value of \(S_n\) can be computed by adding one additional term to the preceding sum, complete Table 8.3.1 with the exact (fractional) value of each sum.
| \(n=1\) | \(S_1 = 1\) |
| \(n=2\) | \(S_2 = \frac{3}{2}\) |
| \(n=3\) | \(S_3 = \frac{7}{4}\) |
| \(n=4\) | \(S_4 = \fillinmath{XXX} \) |
| \(n=5\) | \(S_5 = \fillinmath{XXX} \) |
| \(n=6\) | \(S_6 = \fillinmath{XXX} \) |
| \(n=7\) | \(S_7 = \fillinmath{XXX} \) |
(b)
(c)
Conjecture a simple formula for \(S_n\) that is given by a single fraction where the numerator and/or denominator depend on \(n\text{.}\)
(d)
(e)
How do the results change if we instead consider
\begin{equation*}
S_n = 1 + \frac{1}{4} + \left( \frac{1}{4} \right)^2 + \left( \frac{1}{4} \right)^3 + \cdots + \left( \frac{1}{4} \right)^{n-1}\text{?}
\end{equation*}
