{"id":909,"date":"2025-06-20T17:21:34","date_gmt":"2025-06-20T17:21:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=909"},"modified":"2025-08-28T13:23:40","modified_gmt":"2025-08-28T13:23:40","slug":"sequences-and-series-foundations-background-youll-need-2-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/sequences-and-series-foundations-background-youll-need-2-2\/","title":{"raw":"Power Series and Applications: Background You'll Need 2","rendered":"Power Series and Applications: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the formula for the sum of the first n terms of a geometric series<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Geometric Series<\/h2>\r\nThe sum of the terms in a geometric sequence is called a <strong>geometric series<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>formula for the sum of the first [latex]n[\/latex] terms of a geometric series<\/h3>\r\nA <strong>geometric series<\/strong> is the sum of the terms in a geometric sequence.\r\n\r\nThe formula for the sum of the first [latex]n[\/latex] terms of a geometric sequence is represented as\r\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a geometric series, find the sum of the first [latex]n[\/latex] terms.<\/strong>\r\n<ol>\r\n \t<li>Identify [latex]{a}_{1},r,\\text{ and }n[\/latex].<\/li>\r\n \t<li>Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\r\n \t<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the formula to find the indicated partial sum of each geometric series.\r\n<ol>\r\n \t<li>[latex]{S}_{11}[\/latex] for the series [latex] 8 + -4 + 2 + \\dots [\/latex]\r\n[reveal-answer q=\"373815\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"373815\"][latex]{a}_{1}=8[\/latex], and we are given that [latex]n=11[\/latex]. We can find [latex]r[\/latex] by dividing the second term of the series by the first.\r\n<center>[latex]r=\\dfrac{-4}{8}=-\\frac{1}{2}[\/latex]<\/center>\r\nSubstitute values for [latex]{a}_{1}, r, \\text{ and } n[\/latex] into the formula and simplify.\r\n<center>[latex]\\begin{align}&amp;{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\[1mm] &amp;{S}_{11}=\\dfrac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336 \\\\ \\text{ } \\end{align}[\/latex]<\/center>[\/hidden-answer]<\/li>\r\n \t<li>[latex]\\sum\\limits _{k=1}^6 3\\cdot {2}^{k}[\/latex]\r\n[reveal-answer q=\"369192\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"369192\"]Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.\r\n<center>[latex]{a}_{1}=3\\cdot {2}^{1}=6[\/latex]<\/center>\r\nWe can see from the given explicit formula that [latex]r=2[\/latex]. The upper limit of summation is [latex]6[\/latex], so [latex]n=6[\/latex].Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula, and simplify.<center>[latex]\\begin{align}\\\\ &amp;{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\[1mm] &amp;{S}_{6}=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378 \\end{align}[\/latex]<\/center>[\/hidden-answer]<\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24952[\/ohm2_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24953[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the formula for the sum of the first n terms of a geometric series<\/li>\n<\/ul>\n<\/section>\n<h2>Geometric Series<\/h2>\n<p>The sum of the terms in a geometric sequence is called a <strong>geometric series<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>formula for the sum of the first [latex]n[\/latex] terms of a geometric series<\/h3>\n<p>A <strong>geometric series<\/strong> is the sum of the terms in a geometric sequence.<\/p>\n<p>The formula for the sum of the first [latex]n[\/latex] terms of a geometric sequence is represented as<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a geometric series, find the sum of the first [latex]n[\/latex] terms.<\/strong><\/p>\n<ol>\n<li>Identify [latex]{a}_{1},r,\\text{ and }n[\/latex].<\/li>\n<li>Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\n<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the formula to find the indicated partial sum of each geometric series.<\/p>\n<ol>\n<li>[latex]{S}_{11}[\/latex] for the series [latex]8 + -4 + 2 + \\dots[\/latex]\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q373815\">Show Answer<\/button><\/p>\n<div id=\"q373815\" class=\"hidden-answer\" style=\"display: none\">[latex]{a}_{1}=8[\/latex], and we are given that [latex]n=11[\/latex]. We can find [latex]r[\/latex] by dividing the second term of the series by the first.<\/p>\n<div style=\"text-align: center;\">[latex]r=\\dfrac{-4}{8}=-\\frac{1}{2}[\/latex]<\/div>\n<p>Substitute values for [latex]{a}_{1}, r, \\text{ and } n[\/latex] into the formula and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\[1mm] &{S}_{11}=\\dfrac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336 \\\\ \\text{ } \\end{align}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/li>\n<li>[latex]\\sum\\limits _{k=1}^6 3\\cdot {2}^{k}[\/latex]\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q369192\">Show Answer<\/button><\/p>\n<div id=\"q369192\" class=\"hidden-answer\" style=\"display: none\">Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.<\/p>\n<div style=\"text-align: center;\">[latex]{a}_{1}=3\\cdot {2}^{1}=6[\/latex]<\/div>\n<p>We can see from the given explicit formula that [latex]r=2[\/latex]. The upper limit of summation is [latex]6[\/latex], so [latex]n=6[\/latex].Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula, and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\\\ &{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\[1mm] &{S}_{6}=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378 \\end{align}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24952\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24952&theme=lumen&iframe_resize_id=ohm24952&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24953\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24953&theme=lumen&iframe_resize_id=ohm24953&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":673,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/909"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/909\/revisions"}],"predecessor-version":[{"id":2053,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/909\/revisions\/2053"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/673"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/909\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=909"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=909"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=909"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=909"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}