Geometric Series: Practice

Add the four terms directly.

$3 + 6 + 12 + 24 = 45$

The terms double each time, but for just four of them adding straight across is quickest.

When $r = 1$ the formula divides by $1 - r = 0$, so it cannot be used. But every term equals $a_1$, so you just add $a_1$ to itself $n$ times.

$S_n = \underbrace{a_1 + a_1 + \cdots + a_1}_{n \text{ terms}} = n\,a_1$

Add the six powers of $2$.

$1 + 2 + 4 + 8 + 16 + 32 = 63$

This equals $2^{6} - 1 = 64 - 1 = 63$, one less than the next power of $2$.

Use $S_n = \frac{a_1(r^{n} - 1)}{r - 1}$ with $a_1 = 5$, $r = 2$, $n = 3$.

$S_3 = \frac{5(2^{3} - 1)}{2 - 1} = \frac{5(8 - 1)}{1} = 35$

Directly, $5 + 10 + 20 = 35$.

The first term is $a_1 = \frac12$ and the ratio is $r = \frac12$, with $|r| < 1$, so use $S = \frac{a_1}{1 - r}$.

$S = \frac{1/2}{1 - 1/2} = \frac{1/2}{1/2} = 1$

Divide a term by the one before it. From $1$ to $\frac13$,

$r = \frac{1/3}{1} = \frac13$

The next ratio $\frac{1/9}{1/3} = \frac13$ confirms it.

Every term is $4$, so this is the $r = 1$ case: add $4$ to itself $6$ times.

$S_6 = 6 \times 4 = 24$

With $r = 1$ the sum is $n\,a_1$, not a power.

Divide a term by the one before it.

$r = \frac{4}{8} = \frac12$

The terms are shrinking, so the ratio is a positive fraction less than $1$.

With $a_1 = 8$ and $r = \frac12$ (so $|r| < 1$), use $S = \frac{a_1}{1 - r}$.

$S = \frac{8}{1 - 1/2} = \frac{8}{1/2} = 16$

Add the four terms directly.

$10 + 20 + 40 + 80 = 150$

Add the three terms, which triple at each step.

$5 + 15 + 45 = 65$

The finite geometric sum raises the ratio to the $n$th power, over $1 - r$.

$S_n = \frac{a_1(1 - r^{n})}{1 - r}$

The others are the arithmetic sum $\frac{n(a_1 + a_n)}{2}$, a single term $a_1 r^{n-1}$, and the infinite sum $\frac{a_1}{1 - r}$.