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# 1.5 Finite geometric series

## 1.5 Finite geometric series (EMCDZ)

When we sum a known number of terms in a geometric sequence, we get a finite geometric series. We generate a geometric sequence using the general form:

${T}_{n} = a \cdot {r}^{n-1}$

where

• $$n$$ is the position of the sequence;

• $${T}_{n}$$ is the $$n$$$$^{\text{th}}$$ term of the sequence;

• $$a$$ is the first term;

• $$r$$ is the constant ratio.

### General formula for a finite geometric series (EMCF2)

\begin{align*} {S}_{n} &= a+ ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} \ldots (1) \\ r \times {S}_{n} &= \qquad ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} + ar^{n} \ldots \ldots (2) \\ \text{Subtract eqn. } (2) &\text{ from eqn. } (1) \\ \therefore {S}_{n} - r{S}_{n} &= a + 0 + 0 + \cdots - ar^{n} \\ {S}_{n} - r{S}_{n} &= a - ar^{n} \\ {S}_{n}(1 - r) &= a(1 - r^{n}) \\ \therefore {S}_{n} &= \frac{a(1 - r^{n})}{1 - r} \quad (\text{where } r \ne 1) \end{align*}

The general formula for determining the sum of a geometric series is given by:

${S}_{n} = \frac{a(1 - r^{n})}{1 - r} \qquad \text{where } r \ne 1$

This formula is easier to use when $$r < 1$$.

Alternative formula:

\begin{align*} {S}_{n} &= a+ ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} \ldots \ldots (1) \\ r \times {S}_{n} &= \qquad ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} + ar^{n} \ldots \ldots (2) \\ \text{Subtract eqn. } (1) &\text{ from eqn. } (2) \\ \therefore r{S}_{n} - {S}_{n} &= ar^{n} - a \\ {S}_{n}(r - 1) &= a(r^{n}-1) \\ \therefore {S}_{n} &= \frac{a(r^{n}-1)}{r - 1} \quad (\text{ where } r \ne 1) \end{align*}

The general formula for determining the sum of a geometric series is given by:

${S}_{n} = \frac{a(r^{n}-1)}{r - 1} \qquad \text{where } r \ne 1$

This formula is easier to use when $$r > 1$$.

## Worked example 11: Sum of a geometric series

Calculate: $\sum _{k = 1}^{6}{32 \left( \frac{1}{2} \right)^{k-1}}$

### Write down the first three terms of the series

\begin{align*} k=1; \quad T_{1}&= {32 \left( \frac{1}{2} \right)^{0}} = 32 \\ k=2; \quad T_{2}&= {32 \left( \frac{1}{2} \right)^{2-1}} = 16 \\ k=3; \quad T_{3}&= {32 \left( \frac{1}{2} \right)^{3-1}} = 8 \end{align*}

We have generated the series $$32 + 16 + 8 + \cdots$$

### Determine the values of $$a$$ and $$r$$

\begin{align*} a &= T_{1} = 32 \\ r &= \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} = \frac{1}{2} \end{align*}

### Use the general formula to find the sum of the series

\begin{align*} {S}_{n} &= \frac{a(1 - r^{n})}{1 - r}\\ {S}_{6} &= \frac{32(1 - \left( \frac{1}{2} \right)^{6})}{1 - \frac{1}{2}} \\ &= \frac{32\left(1 - \frac{1}{64} \right)}{\frac{1}{2}} \\ &= 2 \times 32 \left( \frac{63}{64} \right) \\ &= 64 \left( \frac{63}{64} \right) \\ &= 63 \end{align*}

$\sum _{k = 1}^{6}{32 \left( \frac{1}{2} \right)^{k-1}} = 63$
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## Worked example 12: Sum of a geometric series

Given a geometric series with $$T_{1} = -4$$ and $$T_{4} = 32$$. Determine the values of $$r$$ and $$n$$ if $$S_{n} = 84$$.

### Determine the values of $$a$$ and $$r$$

\begin{align*} a &= T_{1} = -4 \\ T_{4} &= ar^{3} = 32 \\ \therefore -4 r^{3} &= 32 \\ r^{3} &= -8 \\ \therefore r &= -2 \end{align*}

Therefore the geometric series is $$-4 + 8 -16 + 32 \ldots$$ Notice that the signs of the terms alternate because $$r < 0$$.

We write the general term for this series as $$T_{n} = -4(-2)^{n-1}$$.

### Use the general formula for the sum of a geometric series to determine the value of $$n$$

\begin{align*} {S}_{n} &= \frac{a(1 - r^{n})}{1 - r}\\ \therefore 84 &= \frac{-4(1 - (-2)^{n})}{1 - (-2)} \\ 84 &= \frac{-4(1 - (-2)^{n})}{3} \\ - \frac{3}{4} \times 84 &= 1 - (-2)^{n} \\ - 63 &= 1 - (-2)^{n} \\ (-2)^{n} &= 64 \\ (-2)^{n} &= (-2)^{6} \\ \therefore n &= 6 \end{align*}

$r = -2 \text{ and } n = 6$
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## Worked example 13: Sum of a geometric series

Use the general formula for the sum of a geometric series to determine $$k$$ if $\sum _{n = 1}^{8}{k \left( \frac{1}{2} \right)^{n}} = \frac{255}{64 }$

### Write down the first three terms of the series

\begin{align*} n=1; \quad T_{1}&= {k \left( \frac{1}{2} \right)^{1}} = \frac{1}{2}k \\ n=2; \quad T_{2}&= {k \left( \frac{1}{2} \right)^{2}} = \frac{1}{4}k \\ n=3; \quad T_{3}&= {k \left( \frac{1}{2} \right)^{3}} = \frac{1}{8}k \end{align*}

We have generated the series $$\frac{1}{2}k + \frac{1}{4}k + \frac{1}{8}k + \cdots$$

We can take out the common factor $$k$$ and write the series as: $$k \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \right)$$

$\therefore k \sum _{n = 1}^{8}{\left( \frac{1}{2} \right)^{n}} = \frac{255}{64}$

### Determine the values of $$a$$ and $$r$$

\begin{align*} a &= T_{1} = \frac{1}{2} \\ r &= \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} = \frac{1}{2} \end{align*}

### Calculate the sum of the first eight terms of the geometric series

\begin{align*} \therefore {S}_{n} &= \frac{a(1 - r^{n})}{1 - r}\\ {S}_{8} &= \frac{\frac{1}{2}(1 - \left( \frac{1}{2} \right)^{8})}{1 - \frac{1}{2}}\\ &= \frac{\frac{1}{2}(1 - \left( \frac{1}{2} \right)^{8})}{\frac{1}{2}}\\ &= 1 - \frac{1}{256} \\ &= \frac{255}{256} \\ & \\ \therefore \sum _{n = 1}^{8}{\left( \frac{1}{2} \right)^{n}} &= \frac{255}{256} \end{align*}

So then we can write:

\begin{align*} k \sum _{n = 1}^{8}{\left( \frac{1}{2} \right)^{n}} &= \frac{255}{64} \\ k \left( \frac{255}{256} \right) &= \frac{255}{64} \\ \therefore k &= \frac{255}{64} \times \frac{256}{255} \\ &= \frac{256}{64} \\ &= 4 \end{align*}

$k = 4$
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## Sum of a geometric series

Textbook Exercise 1.9

Prove that $$a + ar + a{r}^{2} + \cdots + a{r}^{n-1} = \frac{a(r^{n} - 1) }{r - 1}$$ and state any restrictions.

\begin{align*} {S}_{n} &= a+ ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} \ldots (1) \\ r \times {S}_{n} &= \qquad ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} + ar^{n} \ldots (2) \\ \text{Subtract eqn. } (1) &\text{ from eqn. } (2) \\ \therefore r{S}_{n} - {S}_{n} &= ar^{n} - a \\ {S}_{n}(r - 1) &= a(r^{n} - 1) \\ \therefore {S}_{n} &= \frac{a(r^{n} - 1) }{r - 1} \end{align*}

where $$r \ne 1$$.

Given the geometric sequence $$1; -3; 9; \ldots$$ determine:

The eighth term of the sequence.

\begin{align*} a &= 1 \\ r &= \frac{T_{2}}{T_{1}} = -3 \\ T_{n} &= ar^{n-1} \\ \therefore T_{8} &= (1)(-3)^{8-1} \\ &= (1)(-3)^{7} \\ &= -2187 \end{align*}

The sum of the first eight terms of the sequence.

\begin{align*} S_{n} &= \frac{a(1-r^{n})}{1 - r}\\ \therefore S_{8} &= \frac{(1)(1-(-3)^{8})}{1 - (-3)}\\ &= \frac{1 - 6561}{4}\\ &= -\frac{6560}{4}\\ &= -1640 \end{align*}

Determine:

$\sum _{n=1}^{4}3 \cdot {2}^{n-1}$
\begin{align*} S_{4} &= 3 + 6 + 12 + 24 \\ &= 45 \end{align*}

Find the sum of the first $$\text{11}$$ terms of the geometric series $$6+3+\frac{3}{2}+\frac{3}{4}+ \cdots$$

\begin{align*} a &= 6 \\ r &= \frac{1}{2} \\ S_{n} &= \frac{a(1 - r^{n})}{1 - r} \\ S_{11} &= \frac{6(1 - \left( \frac{1}{2} \right)^{11})}{1 - \left( \frac{1}{2} \right)} \\ &= 12 \left( 1 - \frac{1}{2048} \right) \\ &= 12 \left( \frac{2047}{2048} \right) \\ &= \frac{6141}{512} \end{align*}

Show that the sum of the first $$n$$ terms of the geometric series $$54+18+6+\cdots +5 {\left(\frac{1}{3}\right)}^{n-1}$$ is given by $$\left( 81-{3}^{4-n} \right)$$.

\begin{align*} a &= 54 \\ r &= \frac{1}{3} \\ S_{n} &= \frac{a(1 - r^{n})}{1 - r} \\ &= \frac{54(1 - \left( \frac{1}{3} \right)^{n})}{1 - \left( \frac{1}{3} \right)} \\ &= \frac{54(1 - \left( \frac{1}{3} \right)^{n})}{ \frac{2}{3} } \\ &= 81 ( 1 - 3^{-n}) \\ &= 81 - 81 \cdot 3^{-n} \\ &= 81 - (3^{4} \cdot 3^{-n}) \\ &= 81 - 3^{4-n} \end{align*}

The eighth term of a geometric sequence is $$\text{640}$$. The third term is $$\text{20}$$. Find the sum of the first $$\text{7}$$ terms.

\begin{align*} T_{8} &= 640 = ar^{7} \\ T_{3} &= 20 = ar^{2} \\ \therefore \frac{T_{8}}{T_{3}} &= \frac{640}{20} \\ \frac{640}{20} &= \frac{ar^{7}}{ar^{2}} \\ 32 &= r^{5} \\ \therefore 2 &= r \\ \text{And } 20 &= ar^{2} \\ 20 &= a(2)^{2} \\ \frac{20}{4} &= a \\ \therefore 5 &= a \\ r &= 2 \\ S_{n} &= \frac{a(r^{n} - 1)}{r - 1} \\ S_{7} &= \frac{5((2)^{7} - 1)}{2 - 1} \\ &= 5(128 - 1) \\ &= 635 \end{align*}

Given:

$\sum _{t=1}^{n}8 {\left(\frac{1}{2}\right)}^{t}$

Find the first three terms in the series.

\begin{align*} t = 1: \quad T_{1} &= 4 \\ t = 2: \quad T_{2} &= 2 \\ t = 3: \quad T_{3} &= 1 \\ 4; & 2; 1 \end{align*}

Calculate the number of terms in the series if $$S_{n}=7\frac{63}{64}$$.

\begin{align*} a &= 4 \\ r &= \frac{1}{2} \\ S_{n} &= \frac{a(1 - r^{n})}{1 - r} \\ \frac{511}{64} &= \frac{4(1 - \left( \frac{1}{2} \right)^{n})}{1 - \left( \frac{1}{2} \right)} \\ &= \frac{4 - 4\left( \frac{1}{2} \right)^{n}}{\frac{1}{2}} \\ \frac{511}{128} &= 4 - (2^{2} \cdot 2^{-n}) \\ 2^{2 - n}&= 4 - \frac{511}{128} \\ 2^{2 - n}&= \frac{1}{128} \\ 2^{2 - n}&= 2^{-7}\\ 2 - n &= -7 \\ \therefore 9 &= n \end{align*}

The ratio between the sum of the first three terms of a geometric series and the sum of the $$\text{4}$$$$^{\text{th}}$$, $$\text{5}$$$$^{\text{th}}$$ and $$\text{6}$$$$^{\text{th}}$$ terms of the same series is $$8:27$$. Determine the constant ratio and the first $$\text{2}$$ terms if the third term is $$\text{8}$$.

\begin{align*} T_{1} + T_{2} + T_{3} &= a + ar + ar^{2} \\ &= a(1 + r + r^{2}) \\ T_{4} + T_{5} + T_{6} &= ar^{3} + ar^{4} + ar^{5} \\ &= ar^{3}(1 + r + r^{2}) \\ \therefore \frac{T_{1} + T_{2} + T_{3} }{T_{4} + T_{5} + T_{6}} &= \frac{a(1 + r + r^{2})}{ar^{3}(1 + r + r^{2})} \\ \text{And } \quad \frac{T_{1} + T_{2} + T_{3} }{T_{4} + T_{5} + T_{6}} &= \frac{8}{27} \\ \therefore \frac{8}{27} &= \frac{a(1 + r + r^{2})}{ar^{3}(1 + r + r^{2})} \\ &=\frac{1}{r^{3}} \\ \therefore r^{3} &= \frac{27}{8} \\ &= \left( \frac{3}{2} \right)^{3} \\ \therefore r &= \frac{3}{2} \end{align*} \begin{align*} \text{And } T_{3} &= 8 \\ \therefore ar^{2} &= 8 \\ a \left( \frac{3}{2} \right)^{2} &= 8 \\ \therefore a &= 8 \times \frac{4}{9} \\ \therefore T_{1} &= \frac{32}{9} \\ T_{2} &= ar \\ &= \frac{32}{9} \times \frac{3}{2} \\ &= \frac{16}{3} \end{align*}