{"id":1354,"date":"2016-10-26T10:13:44","date_gmt":"2016-10-26T10:13:44","guid":{"rendered":"http:\/\/www.testingdocs.com\/questions\/?p=1354"},"modified":"2024-12-14T04:08:46","modified_gmt":"2024-12-14T04:08:46","slug":"use-mathematical-induction-to-prove-the-statement","status":"publish","type":"post","link":"https:\/\/www.testingdocs.com\/questions\/use-mathematical-induction-to-prove-the-statement\/","title":{"rendered":"Mathematical Induction Example Proof statement"},"content":{"rendered":"<h2>Mathematical Induction Example Proof statement<\/h2>\n<h3>Problem Statement:<\/h3>\n<p>Using <a href=\"https:\/\/www.testingdocs.com\/questions\/what-is-mathematical-induction\/\">Mathematical Induction<\/a>, prove that\u00a0 for n &gt; 1, the following formula holds true:<\/p>\n<p>&nbsp;<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\prod_{i=2}^{n}\\left&amp;space;(&amp;space;1&amp;space;-&amp;space;\\frac{1}{i^{2}}&amp;space;\\right&amp;space;)=&amp;space;\\frac{n+1}{2n}\" alt=\"\\LARGE \\prod_{i=2}^{n}\\left ( 1 - \\frac{1}{i^{2}} \\right )= \\frac{n+1}{2n}\" align=\"absmiddle\" title=\"\"><\/p>\n<h3><\/h3>\n<h3>Mathematical Proof:<\/h3>\n<p>First of all, we will prove for the base case n=2.<\/p>\n<p>LHS\u00a0 =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\left&amp;space;(&amp;space;1&amp;space;-&amp;space;\\frac{1}{2.2}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;\\frac{3}{4}\" alt=\"\\LARGE \\left ( 1 - \\frac{1}{2.2} \\right ) = \\frac{3}{4}\" align=\"absmiddle\" title=\"\"><\/p>\n<p>&nbsp;<\/p>\n<p>RHS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\frac{&amp;space;2&amp;space;+&amp;space;1}{2.2}&amp;space;=&amp;space;\\frac{3}{4}\" alt=\"\\LARGE \\frac{ 2 + 1}{2.2} = \\frac{3}{4}\" align=\"absmiddle\" title=\"\"><\/p>\n<p>&nbsp;<\/p>\n<p>Both LHS and RHS ( Left hand side and Right hand side ) are equal. So its is proved for the base case.<\/p>\n<div>Let us assume that for P(k)\u00a0 for k \u2265 2 be a arbitrarily chosen integer such that the equation holds true.<\/div>\n<div><\/div>\n<div><\/div>\n<div><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\prod_{i=2}^{k}\\left&amp;space;(&amp;space;1&amp;space;-&amp;space;\\frac{1}{i^{2}}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;\\frac{k+1}{2k}\" alt=\"\\LARGE \\prod_{i=2}^{k}\\left ( 1 - \\frac{1}{i^{2}} \\right ) = \\frac{k+1}{2k}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div>Using that we prove that for P(k+1) , the equation holds true, such that :<\/div>\n<div><\/div>\n<div><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\prod_{i=2}^{k+1}\\left&amp;space;(&amp;space;1&amp;space;-&amp;space;\\frac{1}{i^{2}}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;\\frac{(k+1)&amp;space;+1}{2(k+1)}\" alt=\"\\LARGE \\prod_{i=2}^{k+1}\\left ( 1 - \\frac{1}{i^{2}} \\right ) = \\frac{(k+1) +1}{2(k+1)}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div>Let us multiply the k+1 term using the P(k) , we get:<\/div>\n<div><\/div>\n<div><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\prod_{i=2}^{k}\\left&amp;space;(&amp;space;1&amp;space;-&amp;space;\\frac{1}{i^{2}}&amp;space;\\right&amp;space;).\\left&amp;space;(&amp;space;1-\\frac{1}{(k+1)^{2}}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;\\frac{k+1}{2k}.\\left&amp;space;(&amp;space;1-\\frac{1}{(k+1)^{2}}&amp;space;\\right&amp;space;)\" alt=\"\\LARGE \\prod_{i=2}^{k}\\left ( 1 - \\frac{1}{i^{2}} \\right ).\\left ( 1-\\frac{1}{(k+1)^{2}} \\right ) = \\frac{k+1}{2k}.\\left ( 1-\\frac{1}{(k+1)^{2}} \\right )\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>Simplify and get to the form of P(k+1)<\/div>\n<div><\/div>\n<div><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\prod_{i=2}^{k+1}\\left&amp;space;(&amp;space;1&amp;space;-&amp;space;\\frac{1}{i^{2}}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;\\frac{k+1}{2k}.\\left&amp;space;(&amp;space;1-\\frac{1}{(k+1)^{2}}&amp;space;\\right&amp;space;)\" alt=\"\\LARGE \\prod_{i=2}^{k+1}\\left ( 1 - \\frac{1}{i^{2}} \\right ) = \\frac{k+1}{2k}.\\left ( 1-\\frac{1}{(k+1)^{2}} \\right )\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>Expand the RHS and get to the form of p(k+1), we get<\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div>RHS =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\frac{k+1}{2k}&amp;space;-\\frac{k+1}{2k(k+1)^{2}}\" alt=\"\\LARGE \\frac{k+1}{2k} -\\frac{k+1}{2k(k+1)^{2}}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>=\u00a0\u00a0\u00a0\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\frac{k+1}{2k}&amp;space;-&amp;space;\\frac{1}{2k(k+1)}\" alt=\"\\LARGE \\frac{k+1}{2k} - \\frac{1}{2k(k+1)}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>=\u00a0\u00a0\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\frac{(k+1)^2&amp;space;-1}{2k(k+1)}\" alt=\"\\LARGE \\frac{(k+1)^2 -1}{2k(k+1)}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>= \u00a0\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\frac{k^{2}+2k}{2k(k+1)}\" alt=\"\\LARGE \\frac{k^{2}+2k}{2k(k+1)}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>Cancel the k, we get<\/div>\n<div><\/div>\n<div>=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\frac{k+2}{2(k+1)}\" alt=\"\\LARGE \\frac{k+2}{2(k+1)}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\frac{(k+1)+1}{2(k+1)}\" alt=\"\\LARGE \\frac{(k+1)+1}{2(k+1)}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>We proved for the P(k+1) using the P(k)<\/div>\n<div><\/div>\n<div><\/div>\n<div><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;\\prod_{i=2}^{k+1}\\left&amp;space;(&amp;space;1&amp;space;-&amp;space;\\frac{1}{i^{2}}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;\\frac{(k+1)+1}{2(k+1)}\" alt=\"\\LARGE \\prod_{i=2}^{k+1}\\left ( 1 - \\frac{1}{i^{2}} \\right ) = \\frac{(k+1)+1}{2(k+1)}\" align=\"absmiddle\" title=\"\"><\/div>\n<div><\/div>\n<div><\/div>\n<div>Hence proved that:<\/div>\n<div><\/div>\n<div><\/div>\n<div><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\LARGE&amp;space;P(k)&amp;space;\\Rightarrow&amp;space;P(k+1)\" alt=\"\\LARGE P(k) \\Rightarrow P(k+1)\" align=\"absmiddle\" title=\"\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Mathematical Induction Example Proof statement Problem Statement: Using Mathematical Induction, prove that\u00a0 for n &gt; 1, the following formula holds true: &nbsp; Mathematical Proof: First of all, we will prove for the base case n=2. LHS\u00a0 =\u00a0 &nbsp; RHS = &nbsp; Both LHS and RHS ( Left hand side and Right hand side ) are [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[849],"tags":[],"class_list":["post-1354","post","type-post","status-publish","format-standard","hentry","category-math-questions","has-post-title","has-post-date","has-post-category","has-post-tag","has-post-comment","has-post-author",""],"_links":{"self":[{"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/posts\/1354","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/comments?post=1354"}],"version-history":[{"count":11,"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/posts\/1354\/revisions"}],"predecessor-version":[{"id":26382,"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/posts\/1354\/revisions\/26382"}],"wp:attachment":[{"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/media?parent=1354"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/categories?post=1354"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.testingdocs.com\/questions\/wp-json\/wp\/v2\/tags?post=1354"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}