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<div id="main-column" class="span9"> <article id="1c-geometric-sequences-series-sn" style="margin-top: 16px;">
<h1 class="section-title">1C. Geometric sequences & series (SN)</h1>
<ul class="breadcrumb"><li><a title="Home" href="../../../index.html"><i class="fa fa-home"></i></a><span class="divider">/</span></li><li><span class="gray">1. Number & Algebra</span><span class="divider">/</span></li><li><span class="active">1C. Geometric sequences & series (SN)</span></li></ul>
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<h3><img alt="" height="163" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/geo-seq-series-img01.jpg" style="margin-top: 0px; margin-bottom: 0px; float: right;" width="287">∼ Student Notes ∼</h3><h3>Section 1C. Geometric sequences & series</h3><p style="margin-left: 40px;"><strong>Sub-sections:</strong><br><a class="scroll-to" data-target="C1"><u>1C.1</u> General term of a geometric sequence</a><br><a class="scroll-to" data-target="C2"><u>1C.2</u> Graph of a geometric sequence</a><br><a class="scroll-to" data-target="C3"><u>1C.3</u> Geometric series</a><br><a class="scroll-to" data-target="C4"><u>1C.4</u> Summary</a></p><div class="blueBg"><p> <span lang="EN-US" style="font-size:12.0pt;line-height:
115%;font-family:"Times New Roman",serif;mso-fareast-font-family:Calibri;
mso-fareast-theme-font:minor-latin;color:black;mso-themecolor:text1;mso-ansi-language:
EN-US;mso-fareast-language:EN-US;mso-bidi-language:AR-SA">■</span> syllabus content covered in this section ■</p><section class="tib-hiddenbox"><p><strong>SL 1.3* </strong>Geometric sequences and series; use of the formulae for the <span class="math-tex">\(n\)</span><sup>th</sup> term and the sum of the first <em> </em><span class="math-tex">\(n\)</span>terms of the sequence; use of sigma notation for the sums of geometric sequences</p></section></div><hr><h3><a class="anchor" id="C1" name="C1"> </a>1C.1 General term of a geometric sequence</h3><p>Recall the two payment plans, A and B, for a 30-day job described at the start of the previous section (<strong>1B</strong>). We explored the fact that the sequence of daily payments for Plan A is an arithmetic sequence ( <span class="math-tex">\({u_1} = 1\)</span>,<span class="math-tex">\(d = 3\)</span> ). The formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term in an arithmetic series was used to find that the pay earned on the 20<sup>th</sup> day for Plan A is $58. Now, let’s take a closer look at Plan B where you get paid just $0.01 (one cent) for the first day and for each day after that you get paid two times what you were paid the previous day. Plan B's sequence of daily payments for the first week is <span class="math-tex">\(0.01,0.02,0.04,0.08,0.16,0.32,0.64\)</span>. Given the definition from section <strong>1A</strong>, this is clearly a <strong>geometric sequence</strong> where each term is determined by multiplying the previous term by 2; that is, the common ratio, <em>r</em>, is equal to 2. Another way of describing the pattern in this geometric sequence is to say that the <strong>ratio of any term to the previous term</strong> is always 2. For example, <span class="math-tex">\(\frac{{0.32}}{{0.16}} = 2\)</span>, <span class="math-tex">\(\frac{{0.04}}{{0.02}} = 2\)</span>, etc.</p><div class="row-fluid"><div class="span6 col-left"><div><hr class="hidden"><p>The short video at right (no sound) shows how to use the recursive rule for Plan B's sequence (i.e. $0.01 on 1<sup>st</sup> day then multiply by 2 to get pay for each subsequent day) to quickly calculate all 30 terms in the sequence using Excel. Although the amount earned each day is low during the first 10 days or so, the daily pay soon increases rapidly; the amount on the 20<sup>th</sup> day is $5242.88 (compared to $58 for Plan A); and the huge amount of $5,368,709.12 for the 30<sup>th</sup> day! Clearly, you want your employer to use Plan B for your payment plan.</p><hr class="hidden"><p>As we did with arithmetic sequences in the previous section (<strong>1B</strong>), let's find an explicit formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term in a geometric sequence so that we can quickly calculate the value of a term without having to find the value of all the terms before it (method used with Excel in the video).</p></div></div><div class="span6"><div><p style="text-align: center;"><iframe frameborder="0" height="388" scrolling="no" src="https://player.vimeo.com/video/441605794" width="134"></iframe></p></div></div></div><p>Consider a calculation for finding the 5<sup>th</sup> term in Plan B's sequence: <span class="math-tex">\(0.01\left( 2 \right)\left( 2 \right)\left( 2 \right)\left( 2 \right) = 0.01{\left( 2 \right)^4} = 0.16\)</span>. Thus, to find the 5<sup>th</sup>term we multiply the 1<sup>st</sup> term by 2 (common ratio) four times. This makes sense since there are four 'gaps' between the five terms. Hence, for any geometric sequence <span class="math-tex">\({u_5} = {u_1}{r^4}\)</span> (Figure 1); and, the <span class="math-tex">\(n\)</span><sup>th</sup> term can be expressed as <span class="math-tex">\({u_n} = {u_1}{r^{n - 1}}\)</span>. In words, this formula says: "The the <span class="math-tex">\(n\)</span><sup>th</sup> term of a geometric sequence is equal to the 1<sup>st</sup> term multiplied by the common ratio <span class="math-tex">\(\left( {n - 1} \right)\)</span> times."</p><div class="polaroid-center"><img src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/geo-img1.jpg" style="margin:8px 0"><div class="caption">Figure 1</div></div><div class="pinkBg"><h4><span class="math-tex">\(n\)</span><sup>th</sup> term (general term) of a geometric sequence</h4><p>The <span class="math-tex">\(n\)</span><sup>th</sup> term, <span class="math-tex">\({u_n}\)</span>, of a <strong>geometric sequence</strong> with first term <span class="math-tex">\({u_1}\)</span> (initial term) and common ratio <span class="math-tex">\(r\)</span> is given by the following <strong>explicit formula</strong> (in the <a href="/media/ib/mathanalysis/analysis/basics/aa-formula-bklet-v1.2.pdf" target="_blank">Analysis & Approaches formula booklet</a>):</p><p><span class="math-tex">\({u_n} = {u_1}{r^{n - 1}}\)</span></p><p><u>note</u>: Although <span class="math-tex">\(n\)</span> often represents the position (1<sup>st</sup>, 4<sup>th</sup>, etc) of the term in a sequence, it is also common to use other letters such as <span class="math-tex">\(r\)</span> and <span class="math-tex">\(i\)</span> instead of <span class="math-tex">\(n\)</span>. It is also common to use letters other than <span class="math-tex">\(u\)</span> such as <span class="math-tex">\(a\)</span> or <span class="math-tex">\(t\)</span> to represent the terms of a sequence.</p></div><p><strong>Example 1</strong></p><p>Use the formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term of a geometric sequence to show that with payment Plan B (above) you will earn (a) $5242.88 on the 20<sup>th</sup> day, and (b) $5,368,709.12 for the 30<sup>th</sup> day.</p><div class="box"><p><strong>Example 1 solution:</strong></p><section class="tib-hiddenbox"><p>The first term is <span class="math-tex">\({u_1} = 0.01\)</span> and the common ratio is <span class="math-tex">\(r = 2\)</span>; hence, the formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term is <span class="math-tex">\({u_n} = 0.01{\left( 2 \right)^{n - 1}}\)</span></p><p>(a) <span class="math-tex">\({u_{20}} = 0.01{\left( 2 \right)^{20 - 1}} = 0.01{\left( 2 \right)^{19}} = 5242.88\)</span>; thus, you get paid $5242.88 on the 20<sup>th</sup> day</p><p>(b) <span class="math-tex">\({u_{20}} = 0.01{\left( 2 \right)^{30 - 1}} = 0.01{\left( 2 \right)^{29}} = 5368709.12\)</span>; thus, you get paid $5,368,709.12 for the 30<sup>th</sup> day</p></section></div><p><strong>Example 2</strong></p><p>Consider the finite geometric sequence <span class="math-tex">\(4,\;2,\;1,\; \ldots \,,\;\frac{1}{{512}}\)</span>.</p><p>(a) Find the value of the seventh term.</p><p>(b) Find the number of terms in the sequence.</p><div class="box"><p><strong>Example 2 solution:</strong></p><section class="tib-hiddenbox"><p>(a) <span class="math-tex">\({u_1} = 4\)</span> and <span class="math-tex">\(r = \frac{1}{2}\)</span>; hence, <span class="math-tex">\({u_7} = 4{\left( {\frac{1}{2}} \right)^{7 - 1}} = 4{\left( {\frac{1}{2}} \right)^6} = \frac{1}{{16}}\)</span></p><p>(b) Using the formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term, <span class="math-tex">\({u_n} = {u_1}{r^{n - 1}}\)</span>, substitute in the known values: <span class="math-tex">\(\frac{1}{{512}} = 4{\left( {\frac{1}{2}} \right)^{n - 1}}\)</span></p><p>which simplifies to <span class="math-tex">\(\frac{1}{{2048}} = {\left( {\frac{1}{2}} \right)^{n - 1}}\)</span></p><p><span class="math-tex">\(\frac{1}{{2048}} = {\left( {\frac{1}{2}} \right)^{n - 1}}\;\; \Rightarrow \;\;\frac{1}{{2048}} = \frac{1}{{{2^{n - 1}}}}\;\; \Rightarrow \;\;{2^{n - 1}} = 2048\)</span></p><p><span class="math-tex">\({2^n} \cdot {2^{ - 1}} = 2048\;\; \Rightarrow \;\;{2^n} \cdot \frac{1}{2} = 2048\;\; \Rightarrow \;\;{2^n} = 4096\)</span></p><p>At this stage, we could find <span class="math-tex">\(n\)</span> by doing some guess-and-check on our GDC. Some students may even recall that <span class="math-tex">\({2^{10}} = 1024\)</span>.</p><p>Hence, <span class="math-tex">\({2^n} = 4096\;\;\; \Rightarrow \;\;\;n = 12\)</span>. Thus, there are 12 terms in the sequence.</p></section></div><hr class="hidden"><div class="row-fluid"><div class="span6 col-left"><div><p>Now, we have to come up with a strategy to solve for an unknown (<span class="math-tex">\(n\)</span> in this case) that is in the exponent (i.e. solving an exponential equation). Using logarithms is always an effective method for solving an exponential equation, but not always the easiest. Applying some simple algebra is a reasonable way to solve this exponential equation - given that we know <span class="math-tex">\(n\)</span> must be a positive integer since it is the number (position) of the last term.</p></div></div><div class="span6"><div><p style="text-align: center;"><iframe frameborder="0" height="205" scrolling="no" src="https://player.vimeo.com/video/447810624" width="272"></iframe></p></div></div></div><hr class="hidden"><div class="row-fluid"><div class="span6 col-left"><div><p style="text-align: center;"><iframe frameborder="0" height="205" scrolling="no" src="https://player.vimeo.com/video/447815823" width="275"></iframe></p></div></div><div class="span6"><div><p>The video (no sound) above right shows a crude but clear method on the TI84 to show that the sequence has 12 terms (mentally counting while pressing 'enter' to repeatedly multiply by <span class="math-tex">\(\frac{1}{2}\)</span>). Other GDCs, such as the TI-Nspire, do it neater with results automatically displayed as fractions rather than decimals. See TI-Nspire video at left. A more efficient (but perhaps less clear) way of using a GDC to answer Example 2, part (b) solving <span class="math-tex">\(\frac{1}{{512}} = 4{\left( {\frac{1}{2}} \right)^{n - 1}}\)</span> with the GDC's equation solver. This is shown for a TI-Nspire in the image below.</p><p style="text-align: center;"><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/nspire-solve-geo-seq-ex1b-img.jpg" style="margin-top: 0px; margin-bottom: 0px; width: 275px; height: 54px;"></p></div></div></div><hr><h3><a class="anchor" id="C2" name="C2"> </a>1C.2 Graph of a geometric sequence</h3><div class="row-fluid"><div class="span6 col-left"><div><p>As we did for an arithmetic sequence in sub-section 1B.2, we can use a GDC to display a plot (discrete graph) of the terms of a geometric sequence. Let's plot the geometric sequence consisting of the daily payments for Plan B from above, where <span class="math-tex">\({u_1} = 0.01\)</span> and <span class="math-tex">\(r = 2\)</span>. As determined in Example 1, the formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term is <span class="math-tex">\({u_n} = 0.01{\left( 2 \right)^{n - 1}}\)</span>. This is an exponential function where the input, <span class="math-tex">\(n\)</span>, is positive integers from 1 to 30 (# of day) and the output is <span class="math-tex">\({u_n}\)</span> (payment). The video (no sound) at right shows this sequence for Plan B payments being graphed on a TI-Nspire. The graph shows that the points <span class="math-tex">\(\left( {n,{u_n}} \right)\)</span> generated by the formula <span class="math-tex">\({u_n} = 0.01{\left( 2 \right)^{n - 1}}\)</span> lie on a curve that curves sharply upwards starting at about the point where <span class="math-tex">\(n = 26\)</span>. The sequence of payments for Plan B displays rapid, or <strong>exponential</strong>, <strong>growth</strong>.</p><p>A discrete graph for the sequence <span class="math-tex">\(4,\;2,\;1,\; \ldots \,,\;\frac{1}{{512}}\)</span> from Example 2 (video at left) shows that it displays <strong>exponential decay</strong>, i.e. a rapid decrease in the value of the terms. The common ratio for the geometric sequence of Plan B payments is <span class="math-tex">\(r = 2\)</span> yielding exponential growth, and the common ratio for the geometric sequence in Example 2 is <span class="math-tex">\(r = \frac{1}{2}\)</span> yielding exponential decay.</p></div></div><div class="span6"><div><p style="text-align: center;"><iframe frameborder="0" height="244" scrolling="no" src="https://player.vimeo.com/video/448360716" width="325"></iframe></p></div><p style="text-align: center;"><iframe frameborder="0" height="244" scrolling="no" src="https://player.vimeo.com/video/448369670" width="325"></iframe> </p></div></div><p>What happens when the common ratio of a geometric sequence is negative? Consider the geometric sequence where <span class="math-tex">\({u_1} = 1\)</span> and <span class="math-tex">\(r = - \frac{3}{2}\)</span>. The formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term of the sequence is <span class="math-tex">\({u_n} = {\left( { - \frac{3}{2}} \right)^{n - 1}}\)</span>. A discrete graph of the first eight terms of this geometric sequence is shown below in Figure 2.</p><div class="polaroid-right"><img src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/geo-seq-graph-ex2.jpg" style="margin:8px 0"><div class="caption">Figure 2</div></div><p>.</p><p>Clearly, when <span class="math-tex">\(r\)</span> (common ratio) is negative the terms of a geometric sequence <strong>alternate </strong>between positive and negative.</p><div class="greenBg"><h4>Graph (plot) of a geometric sequence</h4><p>The explicit formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term of a <strong>geometric sequence</strong> is an <strong>exponential function</strong> whose plot is a <strong>set of discrete points</strong>, <span class="math-tex">\(\left( {n,{u_n}} \right)\)</span>, that lie on an upward curve (<strong>exponential growth</strong>) if <span class="math-tex">\(r > 1\)</span>, or on a downward curve (<strong>exponential decay</strong>) if <span class="math-tex">\(0 < r < 1\)</span>. If <span class="math-tex">\(r < 0\)</span>, then the signs (<span class="math-tex">\( \pm \)</span>) of the terms alternate and the points do not lie on a single curve.</p></div><hr class="hidden"><hr><h3><a class="anchor" id="C3" name="C3"> </a>1C.3 Geometric series</h3><p>From sub-section 1A.1, we know that the sum of a sequence of numbers is called a series. We need a formula that will compute a geometric series. Computing a series means finding the sum of a sequence but this computation is often referred to as "finding the sum of a sequence", or "finding the sum of a series"; both mean the same thing. In this sub-section, we will only consider geometric series that are <strong>finite</strong>. The sum of an <strong>infinite </strong>geometric series will be covered in Section 1E.</p><p><strong>Example 3</strong></p><p>Find the sum of the series <span class="math-tex">\(1 + 2 + 4 + 8 + \; \ldots \, + 128 + \;256\)</span></p><div class="box"><p><strong>Example 3 solution:</strong></p><section class="tib-hiddenbox"><p>Let <span class="math-tex">\(S\)</span> be the sum of the series: <span class="math-tex">\(S = 1 + 2 + 4 + 8 + \; \ldots \, + 128 + \;256\)</span></p><p>Given that the common ratio is 2, let's multiply both sides by 2: <span class="math-tex">\(2S = 2 + 4 + 8 + \; \ldots \, + \;512\)</span></p><p>and then subtract <span class="math-tex">\(S\)</span> from <span class="math-tex">\(2S\)</span>:</p><p><span class="math-tex">\(\left. {\begin{array}{c}{2S\;\; = \;\;\;\;\;\;\;\;\;\;2 + 4 + 8 + \;\; \ldots \,\; + \;\;512}\\{S\; = 1 + 2 + 4 + \; \ldots \, + \;256}\end{array}} \right\}\;\; \Rightarrow \;\;2S - S = - 1 + 512\;\; \Rightarrow \;\;S = 511\)</span></p></section></div><p>Let's apply the method used in the solution for Example 3 to find the sum of a <em>general </em>geometric series that has <span class="math-tex">\(n\)</span> terms, first term <span class="math-tex">\({u_1}\)</span>, and common ratio <span class="math-tex">\(r\)</span>.</p><p><span class="math-tex">\(\begin{array}{l}{S_n} = {u_1} + {u_1}r + {u_1}{r^2} + \; \cdots \; + {u_1}{r^{n - 1}}\\r{S_n} = \;\;\;\;\;\,{u_1}r + {u_1}{r^2} + \; \cdots \; + {u_1}{r^{n - 1}} + + {u_1}{r^n}\end{array}\)</span></p><p>Now, subtract <span class="math-tex">\(r{S_n}\)</span> (bottom line) from <span class="math-tex">\({S_n}\)</span> (topline), producing: <span class="math-tex">\({S_n} - r{S_n} = {u_1} - {u_1}{r^n}\)</span></p><p>Factorising both sides gives: <span class="math-tex">\({S_n}\left( {1 - r} \right) = {u_1}\left( {1 - {r^n}} \right)\)</span></p><p>Therefore, <span class="math-tex">\({S_n} = \frac{{{u_1}\left( {1 - {r^n}} \right)}}{{1 - r}},\;\;r \ne 1\)</span></p><p><u>Comments</u>: It is possible for <span class="math-tex">\(r = 1\)</span>; for example, <span class="math-tex">\(3 + 3 + 3 + 3 + 3\)</span> is a trivial geometric series. But, in this case, <span class="math-tex">\({S_5} = 5 \cdot 3 = 15\)</span>; and, in general, if <span class="math-tex">\(r = 1\)</span> then <span class="math-tex">\({S_n} = n \cdot {u_1}\)</span>. Also, the formula can be written as <span class="math-tex">\({S_n} = \frac{{{u_1}\left( {{r^n} - 1} \right)}}{{r - 1}}\)</span> (multiplying top and bottom of formula above by <span class="math-tex">\(-1\)</span>). The formula in this form is more convenient when <span class="math-tex">\(r > 1\)</span>.</p><div class="pinkBg"><h4>Sum of a geometric series with <em>n</em> terms</h4><p>The sum of <span class="math-tex">\(n\)</span> terms, <span class="math-tex">\({S_n}\)</span>, of a geometric series with first term <span class="math-tex">\({u_1}\)</span> and common ratio <span class="math-tex">\(r\)</span> is given by the following (in the <a href="/media/ib/mathanalysis/analysis/basics/aa-formula-bklet-v1.2.pdf" target="_blank">Analysis & Approaches formula booklet</a>):</p><p><span class="math-tex">\({S_n} = \frac{{{u_1}\left( {1 - {r^n}} \right)}}{{1 - r}},\;\;r \ne 1\)</span> <span class="math-tex">\(\left[ {{\textrm{or}}\;\;{S_n} = \frac{{{u_1}\left( {{r^n} - 1} \right)}}{{r - 1}}} \right]\)</span></p></div><p><strong>Example 4</strong></p><p>Find the sum of the first five terms of each geometric series.</p><p>(a) <span class="math-tex">\(2 - 6 + 18 - \; \cdots \)</span></p><p>(b) <span class="math-tex">\(6 + 3 + \frac{3}{2} + \; \cdots \)</span></p><p>(c) <span class="math-tex">\(1 + \frac{3}{2} + \frac{9}{4} + \; \cdots \)</span></p><div class="box"><p><strong>Example 4 solution:</strong></p><section class="tib-hiddenbox"><p>(a) <span class="math-tex">\({u_1} = 2\)</span> and <span class="math-tex">\(r = - 3\)</span>: <span class="math-tex">\({S_5} = \frac{{2\left( {1 - {{\left( { - 3} \right)}^5}} \right)}}{{1 - \left( { - 3} \right)}} = \frac{{2\left( {1 - \left( { - 243} \right)} \right)}}{4} = \frac{1}{2}\left( {244} \right) = 122\)</span></p><p>(b) <span class="math-tex">\({u_1} = 6\)</span> and <span class="math-tex">\(r = \frac{1}{2}\)</span>: <span class="math-tex">\({S_5} = \frac{{6\left( {1 - {{\left( {\frac{1}{2}} \right)}^5}} \right)}}{{1 - \frac{1}{2}}} = \frac{{6\left( {1 - \frac{1}{{32}}} \right)}}{{\frac{1}{2}}} = 12\left( {\frac{{31}}{{32}}} \right) = \frac{{93}}{8}\)</span></p><p>(c) <span class="math-tex">\({u_1} = 1\)</span> and <span class="math-tex">\(r = \frac{3}{2}\)</span>: <span class="math-tex">\({S_5} = \frac{{1 \cdot \left( {{{\left( {\frac{3}{2}} \right)}^5} - 1} \right)}}{{\frac{3}{2} - 1}} = \frac{{\frac{{243}}{{32}} - 1}}{{\frac{1}{2}}} = \frac{{\frac{{211}}{{32}}}}{{\frac{1}{2}}} = \frac{{211}}{{32}} \cdot 2 = \frac{{211}}{{16}}\)</span></p></section></div><p>Recall <strong>sigma (<span class="math-tex">\(\Sigma\)</span>) notation</strong> that was covered in sub-sections 1A.1 and 1B.4. The series <span class="math-tex">\(27 + 18 + 12 + 8 + \frac{{16}}{3}\)</span> is geometric such that <span class="math-tex">\({u_1} = 27\)</span> and <span class="math-tex">\(r = \frac{2}{3}\)</span>; hence, <span class="math-tex">\({u_n} = 27{\left( {\frac{2}{3}} \right)^{n - 1}}\)</span>. The most efficient way to express this series is with sigma notation as follows:</p><p><span class="math-tex">\(27 + 18 + 12 + 8 + \frac{{16}}{3} = \sum\limits_{n = 1}^4 {27{{\left( {\frac{2}{3}} \right)}^{n - 1}} = } 27\sum\limits_{n = 1}^4 {{{\left( {\frac{2}{3}} \right)}^{n - 1}}} \)</span></p><p><strong>Example 5</strong></p><p>Evaluate each expression.</p><p>(a) <span class="math-tex">\(\frac{1}{8}\sum\limits_{n = 1}^5 {{2^{n - 1}}} \)</span> (b) <span class="math-tex">\(\sum\limits_{n = 1}^6 {{{\left( { - 2} \right)}^n}} \)</span></p><div class="box"><p><strong>Example 5 solution:</strong></p><section class="tib-hiddenbox"><p>(a) Write out the first three terms of the series.</p><p><span class="math-tex">\(\frac{1}{8}\sum\limits_{n = 1}^5 {{2^{n - 1}}} = \frac{1}{8} \cdot {2^0} + \frac{1}{8} \cdot {2^1} + \frac{1}{8} \cdot {2^2} + \; \cdots \; = \frac{1}{8} + \frac{1}{4} + \frac{1}{2} + \; \cdots \)</span></p><p>It is a geometric series where <span class="math-tex">\({u_1} = \frac{1}{8}\)</span> and <span class="math-tex">\(r = 2\)</span>. Applying the formula for the sum of a geometric series gives</p><p><span class="math-tex">\(\frac{1}{8}\sum\limits_{n = 1}^5 {{2^{n - 1}}} = \frac{{\frac{1}{8}\left( {{2^5} - 1} \right)}}{{2 - 1}} = \frac{1}{8}\left( {32 - 1} \right) = \frac{{31}}{8}\)</span></p><p>(b) Write out the first three terms of the series.</p><p><span class="math-tex">\(\sum\limits_{n = 1}^6 {{{\left( { - \frac{1}{2}} \right)}^n}} = {\left( { - \frac{1}{2}} \right)^1} + {\left( { - \frac{1}{2}} \right)^2} + {\left( { - \frac{1}{2}} \right)^3} + \; \cdots = - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \; \cdots \)</span></p><p>Clearly, this is a geometric series with <span class="math-tex">\({u_1} = - \frac{1}{2}\)</span> and <span class="math-tex">\(r = - \frac{1}{2}\)</span>.</p><p>Thus, <span class="math-tex">\(\sum\limits_{n = 1}^6 {{{\left( { - \frac{1}{2}} \right)}^n}} = \frac{{ - \frac{1}{2}\left( {1 - {{\left( { - \frac{1}{2}} \right)}^6}} \right)}}{{1 - \left( { - \frac{1}{2}} \right)}} = - \frac{1}{3}\left( {1 - \frac{1}{{64}}} \right) = - \frac{1}{3} \cdot \frac{{63}}{{64}} = - \frac{{21}}{{64}}\)</span></p></section></div><p><strong>Example 6</strong></p><p>Given <span class="math-tex">\(9\sum\limits_{r = 1}^k {{{\left( {\frac{1}{3}} \right)}^r} = \frac{{40}}{9}} \)</span>, find the value of <span class="math-tex">\(k\)</span>.</p><div class="box"><p><strong>Example 6 solution:</strong></p><section class="tib-hiddenbox"><p><span class="math-tex">\(9\sum\limits_{r = 1}^k {{{\left( {\frac{1}{3}} \right)}^r} = 3 + } 1 + \frac{1}{3} + \; \cdots \)</span> Hence, it is a geometric series such that <span class="math-tex">\({u_1} = 3\)</span> and <span class="math-tex">\(r = \frac{1}{3}\)</span></p><p>Substituting into the formula for the sum of a geometric series gives</p><p><span class="math-tex">\(\frac{{3\left( {1 - {{\left( {\frac{1}{3}} \right)}^k}} \right)}}{{1 - \frac{1}{3}}} = \frac{{40}}{9}\;\;\; \Rightarrow \;\;\;\frac{9}{2}\left( {1 - {{\left( {\frac{1}{3}} \right)}^k}} \right) = \frac{{40}}{9}\;\;\; \Rightarrow \;\;\;1 - {\left( {\frac{1}{3}} \right)^k} = \frac{{80}}{{81}}\)</span></p><p><span class="math-tex">\({\left( {\frac{1}{3}} \right)^k} = \frac{1}{{81}}\;\;\; \Rightarrow \;\;\;{3^k} = 81\;\;\; \Rightarrow \;\;\;k = 4\)</span></p></section></div><hr><h3><a class="anchor" id="C4" name="C4"> </a>1C.4 Summary</h3><ul><li>The explicit <strong>formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term of a geometric sequence</strong> is: <span class="math-tex">\({u_n} = {u_1}{r^{n - 1}}\)</span></li><li>The explicit <strong>formula </strong>for the <span class="math-tex">\(n\)</span><sup>th</sup> term of a <strong>geometric sequence</strong> is an <strong>exponential function</strong> whose plot is a <strong>set of discrete points</strong>, <span class="math-tex">\(\left( {n,{u_n}} \right)\)</span>, that lie on an upward curve (<span class="math-tex">\(r > 1\)</span>), or a downward curve (<span class="math-tex">\(0 < r < 1\)</span>). If <span class="math-tex">\(r < 0\)</span>, then the terms alternate signs (±) and do not lie on a single curve.</li><li>The terms in a geometric sequence can display <strong>exponential growth</strong> (<span class="math-tex">\(r > 1\)</span>) or <strong>exponential decay</strong> (<span class="math-tex">\(0 < r < 1\)</span>).</li><li>The <strong>formula to calculate the sum of a geometric series</strong> is: <span class="math-tex">\({S_n} = \frac{{{u_1}\left( {1 - {r^n}} \right)}}{{1 - r}},\;\;r \ne 1\)</span> <span class="math-tex">\(\left[ {{\textrm{or}}\;\;{S_n} = \frac{{{u_1}\left( {{r^n} - 1} \right)}}{{r - 1}}} \right]\)</span></li><li><strong>Sigma (<span class="math-tex">\(\Sigma\)</span>) notation</strong> is a convenient and efficient way to represent geometric series. <span class="math-tex">\(\sum\limits_{r = 1}^n {{u_r}} \)</span> represents the sum of <span class="math-tex">\(n\)</span> terms where <span class="math-tex">\({{u_r}}\)</span> is the algebraic rule (an exponential function) for generating each term.</li></ul><hr><table width="100%"><tbody><tr><td>← go to previous section: <a href="../34896/1b-arithmetic-sequences-series-sn.html" target="_self">1B. Arithmetic sequences & series</a></td><td style="text-align: right;"> go to next section: 1D. Infinite geometric series → </td></tr></tbody></table><script>document.querySelectorAll('.tib-teacher-only').forEach(e => e.remove());</script>
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