Prove that \(a + ar + a{r}^{2} + \cdots + a{r}^{n-1} = \frac{a(r^{n} - 1) }{r - 1}\) and state any restrictions.
where \(r \ne 1\).
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1.4 Finite arithmetic series
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1.6 Infinite series
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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.
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 \(r > 1\).
Calculate: \[\sum _{k = 1}^{6}{32 \left( \frac{1}{2} \right)^{k-1}}\]
We have generated the series \(32 + 16 + 8 + \cdots\)
Given a geometric series with \(T_{1} = -4\) and \(T_{4} = 32\). Determine the values of \(r\) and \(n\) if \(S_{n} = 84\).
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 \(k\) if \[\sum _{n = 1}^{8}{k \left( \frac{1}{2} \right)^{n}} = \frac{255}{64 }\]
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: \(k \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \right)\)
\[\therefore k \sum _{n = 1}^{8}{\left( \frac{1}{2} \right)^{n}} = \frac{255}{64}\]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*}Prove that \(a + ar + a{r}^{2} + \cdots + a{r}^{n-1} = \frac{a(r^{n} - 1) }{r - 1}\) and state any restrictions.
where \(r \ne 1\).
Given the geometric sequence \(1; -3; 9; \ldots\) determine:
The eighth term of the sequence.
The sum of the first eight terms of the sequence.
Determine:
\[\sum _{n=1}^{4}3 \cdot {2}^{n-1}\]Find the sum of the first \(\text{11}\) terms of the geometric series \(6+3+\frac{3}{2}+\frac{3}{4}+ \cdots\)
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 \(\left( 81-{3}^{4-n} \right)\).
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.
Given:
\[\sum _{t=1}^{n}8 {\left(\frac{1}{2}\right)}^{t}\]Find the first three terms in the series.
Calculate the number of terms in the series if \(S_{n}=7\frac{63}{64}\).
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 \(\text{8}\).
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1.4 Finite arithmetic series
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1.6 Infinite series
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