{"id":104918,"date":"2021-01-30T14:21:27","date_gmt":"2021-01-30T08:51:27","guid":{"rendered":"https:\/\/www.vskills.in\/certification\/tutorial\/?page_id=104918"},"modified":"2024-04-12T14:28:18","modified_gmt":"2024-04-12T08:58:18","slug":"mathematical-derivation-of-formula-for-future-compound-value-of-annuity","status":"publish","type":"page","link":"https:\/\/www.vskills.in\/certification\/tutorial\/mathematical-derivation-of-formula-for-future-compound-value-of-annuity\/","title":{"rendered":"Mathematical Derivation of Formula for Future\/Compound Value of Annuity"},"content":{"rendered":"\n<p>Fan = A (1 + r)n-1+ A (1 + r) n- +\u2026\u2026\u2026\u2026<\/p>\n\n\n\n<p>\u2026\u2026\u2026&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..\u2026\u2026+ A (1 + r)1 + A (1 + r)0<\/p>\n\n\n\n<p>When we take out A in every item, the series is as below:<\/p>\n\n\n\n<p>(1 + r)n-1 + (1 + r) n- +\u2026\u2026\u2026\u2026\u2026<\/p>\n\n\n\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..\u2026\u2026\u2026\u2026+ (1 + r)1 + (1 + r)0<\/p>\n\n\n\n<p>We may rewrite the series as below also:<\/p>\n\n\n\n<p>(1 + r)0 + (1 + r)1 + \u2026\u2026\u2026\u2026\u2026\u2026\u2026 .<\/p>\n\n\n\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;\u2026.+ (1 + r) n-2 + (1 + r)n-1<\/p>\n\n\n\n<p>This series of n terms is in Geometric Progression with a common ratio. The Common Ratio is arrived at by dividing the following term by the preceding term in the series.<\/p>\n\n\n\n<p>Let us take the first term (1 + r)0 which is 1 (anything raised to the power of 0 is equal to 1). Dividing the second term by the first term 1, we get (1 + r) i.e. (1 + r)1 \/ 1. Next we shall divide the third item (1 + r)2 by (1 + r)1 which is again (1 + r). Hence the Common Ratio in our series is (1 + r).<\/p>\n\n\n\n<p>We get Son by adding the above n number of terms in the series<\/p>\n\n\n\n<p>Son = (1 + r)<sup>0<\/sup> + (1 + r)<sup>1<\/sup> +.\u2026.+ (1 + r) <sup>n-2<\/sup> + (1 + r)<sup>n-1<\/sup> \u2026\u2026.Eqn.1<\/p>\n\n\n\n<p>Multiplying both sides by \u2018(1 + r)\u2019 we have,<\/p>\n\n\n\n<p>(1+r)S<sub>on<\/sub> = (1+r) (1 + r)<sup>0<\/sup> + (1+r) (1 + r)<sup>1<\/sup> + \u2026\u2026\u2026\u2026\u2026&#8230;&#8230;<\/p>\n\n\n\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;. + (1+r) (1 + r) <sup>n-2 <\/sup>&nbsp;+ (1+r) (1 + r)<sup>n-1<\/sup><\/p>\n\n\n\n<p>(1+r)Son = (1+r)&nbsp; +&nbsp; (1 + r)<sup>2<\/sup> + \u2026. &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..<\/p>\n\n\n\n<p>+&nbsp; (1 + r) <sup>n-1&nbsp; <\/sup>+&nbsp; (1 + r)<sup>n <\/sup>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.\u2026Eqn. 2<\/p>\n\n\n\n<p>Now Subtracting Equation 2 from Equation 1, we have\\<\/p>\n\n\n\n<p>Son \u2013 (1 + r)Sn = 1 &#8211;&nbsp; (1 + r)<sup>n<\/sup><\/p>\n\n\n\n<p>Sn [1- (1+r)] = 1 &#8211; (1 + r)<sup>n<\/sup><\/p>\n\n\n\n<p>Multiplying with A which was removed in the beginning, the formula is<\/p>\n\n\n\n<p>A {[ (1 + r)<sup>n<\/sup> \u2013 1] \/ r}<\/p>\n\n\n\n<p>Mathematical Derivation of formula for Present Value of an<\/p>\n\n\n\n<p><strong>Annuity<\/strong><\/p>\n\n\n\n<p>PVAn = A [1 \/ (1 + r)<sup>1<\/sup>] + A[ 1 \/ (1 + r)<sup>2<\/sup>] + \u2026\u2026\u2026\u2026\u2026<\/p>\n\n\n\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.\u2026.+ A[1 \/ (1 + r)n]\n\n\n\n<p>We may rewrite the equation as follows:<\/p>\n\n\n\n<p>PVAn = A [ (1 + r)<sup>-1<\/sup>] + A[ (1 + r)<sup>-2<\/sup>] + \u2026\u2026\u2026\u2026\u2026<\/p>\n\n\n\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.\u2026.+ A[(1 + r)-n]\n\n\n\n<p>When we take out A in every item, the series is as below:<\/p>\n\n\n\n<p>(1 + r)<sup>-1<\/sup> + (1 + r)-<sup>2<\/sup> + \u2026\u2026\u2026\u2026\u2026\u2026.+ (1 + r)-<sup>n<\/sup><\/p>\n\n\n\n<p>This series of n terms is in Geometric Progression with a common ratio.&nbsp; The Common Ratio is arrived at by dividing the following term by the preceding term in the series. Dividing the second term by the first term, we get (1 + r)-1. Next we shall divide the third item (1 + r)-3 by (1 + r)-2 which is again (1 + r)-1.&nbsp; Hence the Common Ratio in our series is (1 + r)- 1.<\/p>\n\n\n\n<p>We get Sn by adding the above n number of terms in the series<\/p>\n\n\n\n<p>Sn = (1 + r)-<sup>1<\/sup> + (1 + r)-<sup>2<\/sup> + \u2026\u2026\u2026\u2026.+ (1 + r)-<sup>n<\/sup> \u2026&#8230;&#8230;Eqn.1<\/p>\n\n\n\n<p>Multiplying both sides by \u2018(1 +&nbsp; r)-1\u2019 we have,<\/p>\n\n\n\n<p>(1+r)-1Sn = (1 + r)-<sup>1<\/sup>(1 + r)-1&nbsp; + (1 + r)-<sup>2<\/sup> (1 + r)-1 + \u2026\u2026\u2026\u2026\u2026<\/p>\n\n\n\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..\u2026.+ (1 + r)<sup>-n<\/sup>(1 + r)<sup>-1<\/sup><\/p>\n\n\n\n<p>Using the concept of&nbsp; am X an = am+n, we get<\/p>\n\n\n\n<p>(1+r)-1Sn = (1 + r)<sup>-2<\/sup> + (1 + r)<sup>-3<\/sup> + \u2026\u2026\u2026\u2026\u2026\u2026.+ (1 + r<sup>)-n-1<\/sup><\/p>\n\n\n\n<p>\u2026\u2026\u2026\u2026Eqn. 2<\/p>\n\n\n\n<p>Now Subtracting Equation 2 from Equation 1, we have<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"http:\/\/www.vskills.in\/lms\/wp-content\/uploads\/2016\/06\/Image-30-5.jpg\" alt=\"Image 30\" class=\"wp-image-39886\"\/><\/figure><\/div>\n\n\n\n<p>Simplifying the denominator,<\/p>\n\n\n\n<p>1- (1+r)-1 =&nbsp;&nbsp; 1 &#8211;&nbsp;&nbsp;&nbsp; 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=&nbsp;&nbsp; &nbsp;&nbsp;1+ r \u2013 1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;=&nbsp;&nbsp; r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;<\/p>\n\n\n\n<p>1 + r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;1 + r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;1 + r<\/p>\n\n\n\n<p>Simplifying the numerator,<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"http:\/\/www.vskills.in\/lms\/wp-content\/uploads\/2016\/06\/Image-31-5.jpg\" alt=\"Image 31\" class=\"wp-image-39887\"\/><\/figure><\/div>\n\n\n\n<p>Using the simplified numerator and denominator,<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"http:\/\/www.vskills.in\/lms\/wp-content\/uploads\/2016\/06\/Image-32-3.jpg\" alt=\"Image 32\" class=\"wp-image-39892\"\/><\/figure><\/div>\n\n\n\n<p>Multiplying with A which was removed in the beginning, the formula is<\/p>\n\n\n\n<p>A {[ (1 + r)n \u2013 1] \/ [(1+r)n&nbsp; X&nbsp; r]}<\/p>\n\n\n\n<p>Mathematical Derivation of formula for Present Value of an Annuity<\/p>\n\n\n\n<p>PVADn = A {1+[1 \/ (1 + r)1] + [1 \/ (1 + r)2] +<\/p>\n\n\n\n<p>\u2026\u2026\u2026\u2026\u2026\u2026.+ [1 \/ (1+r)n-1] }<\/p>\n\n\n\n<p>Multiplying and dividing the whole expression by (1+r), we get<\/p>\n\n\n\n<p>A (1+r)&nbsp; {1+[1 \/ (1 + r)1] + [1 \/ (1 + r)2] + \u2026\u2026\u2026\u2026\u2026\u2026.+ [1 \/ (1+r)n-1] }<\/p>\n\n\n\n<p>&#8212;&#8212;-<\/p>\n\n\n\n<p>(1+r)<\/p>\n\n\n\n<p>A (1+r)&nbsp; {1+[1 \/ (1 + r)1] + [1 \/ (1 + r)2] + \u2026\u2026\u2026\u2026\u2026\u2026.+ [1 \/ (1+r)n-1] } x [1\/(1+r)]\n\n\n\n<p>A (1+r) {[1 \/ (1 + r)] + [1 \/ (1 + r)2] + \u2026\u2026\u2026\u2026\u2026\u2026.+ [1 \/ (1+r)n] }<\/p>\n\n\n\n<p>Keeping the term A(1+r), the other term is in geometric progression with common ratio as 1\/(1+r).&nbsp; Hence applying the formula for sum of geometric progression i.e. Sn = a(rn \u2013 1) \/ r \u20131 where a is the first term and r is the common ratio, we get<\/p>\n\n\n\n<p>(1 + r)n \u2013 1 \/ [(1+r)n&nbsp; X&nbsp; r<\/p>\n\n\n\n<p>Adding the term already removed we have A (1+r) [(1 + r)n \u2013 1<\/p>\n\n\n\n<p>\/ [(1+r)n&nbsp; X&nbsp; r]\n","protected":false},"excerpt":{"rendered":"<p>Fan = A (1 + r)n-1+ A (1 + r) n- +\u2026\u2026\u2026\u2026 \u2026\u2026\u2026&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..\u2026\u2026+ A (1 + r)1 + A (1 + r)0 When we take out A in every item, the series is as below: (1 + r)n-1 + (1 + r) n- +\u2026\u2026\u2026\u2026\u2026 &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..\u2026\u2026\u2026\u2026+ (1 + r)1 + (1 + r)0 We may rewrite&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-104918","page","type-page","status-publish","hentry"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v24.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Mathematical Derivation of Formula for Future\/Compound Value of Annuity - Tutorial<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.vskills.in\/certification\/tutorial\/mathematical-derivation-of-formula-for-future-compound-value-of-annuity\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Mathematical Derivation of Formula for Future\/Compound Value of Annuity - Tutorial\" \/>\n<meta property=\"og:description\" content=\"Fan = A (1 + r)n-1+ A (1 + r) n- +\u2026\u2026\u2026\u2026 \u2026\u2026\u2026&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..\u2026\u2026+ A (1 + r)1 + A (1 + r)0 When we take out A in every item, the series is as below: (1 + r)n-1 + (1 + r) n- +\u2026\u2026\u2026\u2026\u2026 &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..\u2026\u2026\u2026\u2026+ (1 + r)1 + (1 + r)0 We may rewrite...\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.vskills.in\/certification\/tutorial\/mathematical-derivation-of-formula-for-future-compound-value-of-annuity\/\" \/>\n<meta property=\"og:site_name\" content=\"Tutorial\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/vskills.in\/\" \/>\n<meta property=\"article:modified_time\" content=\"2024-04-12T08:58:18+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/www.vskills.in\/lms\/wp-content\/uploads\/2016\/06\/Image-30-5.jpg\" \/>\n<meta name=\"twitter:label1\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data1\" content=\"3 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.vskills.in\/certification\/tutorial\/mathematical-derivation-of-formula-for-future-compound-value-of-annuity\/\",\"url\":\"https:\/\/www.vskills.in\/certification\/tutorial\/mathematical-derivation-of-formula-for-future-compound-value-of-annuity\/\",\"name\":\"Mathematical Derivation of Formula for Future\/Compound Value of Annuity - 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