Geometric Series: Practice

With $a_1 = 2$, $r = 3$, $n = 5$,

$S_5 = \frac{2(3^{5} - 1)}{3 - 1} = \frac{2(243 - 1)}{2} = 242$

The summand $3 \cdot 2^{k-1}$ gives the terms $3, 6, 12, 24$ as $k$ runs from $1$ to $4$, a geometric series with $a_1 = 3$, $r = 2$.

$\sum_{k=1}^{4} 3 \cdot 2^{k-1} = \frac{3(2^{4} - 1)}{2 - 1} = 3(15) = 45$

Here $a_1 = 9$ and $r = \frac13$, with $|r| < 1$, so $S = \frac{a_1}{1 - r}$.

$S = \frac{9}{1 - 1/3} = \frac{9}{2/3} = \frac{27}{2}$

Write the decimal as a series: $0.\overline{5} = \frac{5}{10} + \frac{5}{100} + \cdots$, with $a_1 = \frac{5}{10}$ and $r = \frac{1}{10}$.

$0.\overline{5} = \frac{5/10}{1 - 1/10} = \frac{5/10}{9/10} = \frac59$

With $a_1 = \frac{7}{10}$ and $r = \frac{1}{10}$,

$0.\overline{7} = \frac{7/10}{1 - 1/10} = \frac{7/10}{9/10} = \frac79$

With $a_1 = 1$, $r = 2$, $n = 10$,

$S_{10} = \frac{1(2^{10} - 1)}{2 - 1} = 1024 - 1 = 1023$

With $a_1 = 4$, $r = 3$, $n = 5$,

$S_5 = \frac{4(3^{5} - 1)}{3 - 1} = \frac{4(242)}{2} = 4 \times 121 = 484$

Here $a_1 = \frac34$ and $r = \frac14$, with $|r| < 1$.

$S = \frac{3/4}{1 - 1/4} = \frac{3/4}{3/4} = 1$

This is a finite sum with $a_1 = \frac12$, $r = \frac12$, $n = 4$.

$S_4 = \frac{\frac12\left(1 - (1/2)^{4}\right)}{1 - 1/2} = 1 - \left(\tfrac12\right)^{4} = \frac{15}{16}$

Adding $\frac{8}{16} + \frac{4}{16} + \frac{2}{16} + \frac{1}{16}$ gives the same $\frac{15}{16}$.

The three terms model three years of ten percent growth: $100$, $110$, $121$.

$100 + 110 + 121 = 331$

A two-digit repeat has $a_1 = \frac{27}{100}$ and $r = \frac{1}{100}$.

$0.\overline{27} = \frac{27/100}{1 - 1/100} = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}$

This series has $a_1 = 2$, $r = 3$, $n = 6$.

$S_6 = \frac{2(3^{6} - 1)}{3 - 1} = \frac{2(728)}{2} = 728$