General Formula For a Finite Geometric Series

\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} &= \cfrac{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:

\[{S}_{n} = \cfrac{a(1 - r^{n})}{1 - r} \qquad \text{where } r \ne 1\]

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} &= \cfrac{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:

\[{S}_{n} = \cfrac{a(r^{n}-1)}{r - 1} \qquad \text{where } r \ne 1\]

This formula is easier to use when \(r > 1\).

Example

Question

Calculate: \[\sum _{k = 1}^{6}{32 ( \cfrac{1}{2} )^{k-1}}\]

Write down the first three terms of the series

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

Use the general formula to find the sum of the series

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

Write the final answer

\[\sum _{k = 1}^{6}{32 ( \cfrac{1}{2} )^{k-1}} = 63\]

Example

Question

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

Write the final answer

\[r = -2 \text{ and } n = 6\]

Example

Question

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

Write down the first three terms of the series

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

We have generated the series \(\cfrac{1}{2}k + \cfrac{1}{4}k + \cfrac{1}{8}k + \cdots\)

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

\[\therefore k \sum _{n = 1}^{8}{( \cfrac{1}{2} )^{n}} = \cfrac{255}{64}\]

Determine the values of \(a\) and \(r\)

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

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

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

So then we can write:

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