{"id":1514,"date":"2021-05-09T15:35:42","date_gmt":"2021-05-09T19:35:42","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=1514"},"modified":"2024-10-25T02:48:54","modified_gmt":"2024-10-25T06:48:54","slug":"math-olympiad-exercise-1","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=1514","title":{"rendered":"Math Olympiad Exercise &#8211; 1"},"content":{"rendered":"\n<p>Find the formula for the following summary: $$\\sum_{k=1}^{n}(\\dfrac{1}{2k}-\\dfrac{1}{2k+1}+\\dfrac{1}{k+n})$$<\/p>\n\n\n\nClick <a onclick=\"toggle_visibility('math-olympiad-exercise-1');\">here<\/a> for the solution.\n<div id=\"math-olympiad-exercise-1\" style=\"display:none\">\n\n\n\n<p><strong>Solution<\/strong>: The formula is $$ S(n)=\\sum_{k=1}^{n}(\\dfrac{1}{2k}-\\dfrac{1}{2k+1}+\\dfrac{1}{k+n})=\\dfrac{2n}{2n+1}$$<\/p>\n\n\n\n<p><em>Proof by induction<\/em>: For $n=1$, $$S(1)=\\dfrac{1}{2}-\\dfrac{1}{3}+\\dfrac{1}{2}=\\dfrac{2}{3}=\\dfrac{2\\cdot 1}{2\\cdot 1+1}$$ and the claim is correct. <\/p>\n\n\n\n<p>Assume for $n=m$, the claim is correct, therefore $$S(m)=\\dfrac{2m}{2m+1}$$ Then $$ S(m+1)=S(m)-\\dfrac{1}{m+1}+(\\dfrac{1}{2(m+1)}-\\dfrac{1}{2(m+1)+1}+\\dfrac{1}{2(m+1)})+\\dfrac{1}{2m+1}$$ $$ = S(m)-\\dfrac{1}{2m+3}+\\dfrac{1}{2m+1}=\\dfrac{2m}{2m+1}+\\dfrac{2}{(2m+1)(2m+3)}$$ $$ = \\dfrac{2m(2m+3)+2}{(2m+1)(2m+3)}=\\dfrac{2(m+1)(2m+1)}{(2m+1)(2m+3)}=\\dfrac{2(m+1)}{2(m+1)+1}$$<\/p>\n\n\n\n<p>Therefore, for $n=m+1$, the claim is correct, and the formula is proved by induction.<\/p>\n\n\n\n<p><em>Proof by expansion<\/em>: By expanding the summary, it can be shown that all fractions are cancelled out, except the following $$ \\dfrac{1}{2}+\\dfrac{1}{2^2}+\\dfrac{1}{2^3}+&#8230;+\\dfrac{1}{2^m}+\\dfrac{1}{2^m}-\\dfrac{1}{2n+1}=1-\\dfrac{1}{2n+1}=\\dfrac{2n}{2n+1}$$ where $2^m$ is the largest power of 2 less or equal to $n$.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Find the formula for the following summary: $$\\sum_{k=1}^{n}(\\dfrac{1}{2k}-\\dfrac{1}{2k+1}+\\dfrac{1}{k+n})$$ Click here for the solution. Solution: The formula is $$ S(n)=\\sum_{k=1}^{n}(\\dfrac{1}{2k}-\\dfrac{1}{2k+1}+\\dfrac{1}{k+n})=\\dfrac{2n}{2n+1}$$ Proof by induction: For $n=1$, $$S(1)=\\dfrac{1}{2}-\\dfrac{1}{3}+\\dfrac{1}{2}=\\dfrac{2}{3}=\\dfrac{2\\cdot 1}{2\\cdot 1+1}$$ and the claim is correct. Assume for $n=m$, the claim is correct, therefore &hellip; <a href=\"https:\/\/mathfun4kids.com\/mlog\/?p=1514\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1514"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1514"}],"version-history":[{"count":27,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1514\/revisions"}],"predecessor-version":[{"id":4571,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1514\/revisions\/4571"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1514"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1514"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1514"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}