{"id":1539,"date":"2021-05-09T18:34:14","date_gmt":"2021-05-09T18:34:14","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=1539"},"modified":"2024-10-25T08:59:33","modified_gmt":"2024-10-25T12:59:33","slug":"lmt-2021-team-round-problem-17","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=1539","title":{"rendered":"LMT 2021 Team Round &#8211; Problem 17"},"content":{"rendered":"\n<p>Given that the value of $$\\sum_{k=1}^{2021}\\dfrac{1}{1^2+2^2+3^2+&#8230;+k^2} + \\sum_{k=1}^{1010}\\dfrac{6}{2k^2-k} + \\sum_{k=1011}^{2021}\\dfrac{24}{2k+1}$$ can be expressed as $\\dfrac{n}{m}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.<\/p>\n\n\n\nClick <a onclick=\"toggle_visibility('lmt-2021-team-17');\">here<\/a> for the solution.\n<div id=\"lmt-2021-team-17\" style=\"display:none\">\n\n\n\n<p><strong>Solution<\/strong>: $$S=\\sum_{k=1}^{2021}\\dfrac{1}{1^2+2^2+3^2+&#8230;+k^2} + \\sum_{k=1}^{1010}\\dfrac{6}{2k^2-k} + \\sum_{k=1011}^{2021}\\dfrac{24}{2k+1}$$ $$=\\sum_{k=1}^{2021}\\dfrac{6}{k(k+1)(2k+1)} + \\sum_{k=1}^{1010}\\dfrac{6}{k(2k-1)} + \\sum_{k=1011}^{2021}\\dfrac{24}{2k+1}$$ $$=\\sum_{k=1}^{2021}\\dfrac{6}{k(k+1)(2k+1)} + (6 + \\sum_{k=1}^{1010}\\dfrac{6}{(k+1)(2k+1)}-\\dfrac{6}{1011\\cdot 2021} )+ \\sum_{k=1011}^{2021}\\dfrac{24}{2k+1}$$ <\/p>\n\n\n\n<p>Let $$C=1-\\dfrac{1}{1011\\cdot 2021}=\\dfrac{1011\\cdot 2021-1}{1011\\cdot 2021}$$ We have:<\/p>\n\n\n\n<p>$$S=6\\cdot(\\sum_{k=1}^{1010}(\\dfrac{1}{k(k+1)(2k+1)}+\\dfrac{1}{(k+1)(2k+1)}) + \\sum_{k=1011}^{2011}(\\dfrac{1}{k(k+1)(2k+1)} +\\dfrac{4}{2k+1}) + C)$$ $$=6\\cdot(\\sum_{k=1}^{1010}\\dfrac{k+1}{k(k+1)(2k+1)}+\\sum_{k=1011}^{2021}\\dfrac{1+4k(k+1)}{k(k+1)(2k+1)} + C)$$ $$=6\\cdot(\\sum_{k=1}^{1010}\\dfrac{1}{k(2k+1)} + \\sum_{k=1011}^{2021}\\dfrac{2k+1}{k(k+1)} + C)$$ $$=6\\cdot(\\sum_{k=1}^{1010}(\\dfrac{2}{2k}-\\dfrac{2}{2k+1}) +\\sum_{k=1011}^{2021}(\\dfrac{1}{k}+\\dfrac{1}{k+1}) +C$$ $$=6\\cdot(\\sum_{k=1}^{1010}(\\dfrac{2}{2k}-\\dfrac{2}{2k+1}) +\\sum_{k=1}^{1011}(\\dfrac{1}{k+1010}+\\dfrac{1}{k+1011}) + C)$$ $$=6\\cdot(\\sum_{k=1}^{1010}(\\dfrac{2}{2k}-\\dfrac{2}{2k+1}) +(-\\dfrac{1}{1011}+\\sum_{k=1}^{1011}\\dfrac{2}{k+1010}+\\dfrac{1}{2022}) + C)$$ $$=6\\cdot(2\\cdot\\sum_{k=1}^{1010}(\\dfrac{1}{2k}-\\dfrac{1}{2k+1}+\\dfrac{1}{k+1010})-\\dfrac{1}{2022}+\\dfrac{2}{2021}+C)$$ <\/p>\n\n\n\n<p>Because (<a href=\"https:\/\/mathfun4kids.com\/mlog\/?p=1514\">https:\/\/mathfun4kids.com\/mlog\/?p=1514<\/a>) $$\\sum_{k=1}^{n}(\\dfrac{1}{2k}-\\dfrac{1}{2k+1}+\\dfrac{1}{k+n})=\\dfrac{2n}{2n+1}$$ Therefore $$S=6\\cdot(2\\cdot \\dfrac{2\\cdot 1010}{2\\cdot 1010 + 1}-\\dfrac{1}{2022}+\\dfrac{2}{2021}+\\dfrac{1011\\cdot 2021-1}{1011\\cdot 2021})$$ $$=6\\cdot(2-\\dfrac{1}{2022} + \\dfrac{1011\\cdot 2021-1}{1011\\cdot 2021}) = \\dfrac{12257363}{681077}$$<\/p>\n\n\n\n<p>Therefore, $m+n=681077+12257463=\\boxed{12938440}$<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Given that the value of $$\\sum_{k=1}^{2021}\\dfrac{1}{1^2+2^2+3^2+&#8230;+k^2} + \\sum_{k=1}^{1010}\\dfrac{6}{2k^2-k} + \\sum_{k=1011}^{2021}\\dfrac{24}{2k+1}$$ can be expressed as $\\dfrac{n}{m}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$. Click here for the solution. Solution: $$S=\\sum_{k=1}^{2021}\\dfrac{1}{1^2+2^2+3^2+&#8230;+k^2} + \\sum_{k=1}^{1010}\\dfrac{6}{2k^2-k} + \\sum_{k=1011}^{2021}\\dfrac{24}{2k+1}$$ $$=\\sum_{k=1}^{2021}\\dfrac{6}{k(k+1)(2k+1)} + \\sum_{k=1}^{1010}\\dfrac{6}{k(2k-1)} &hellip; <a href=\"https:\/\/mathfun4kids.com\/mlog\/?p=1539\">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\/1539"}],"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=1539"}],"version-history":[{"count":50,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1539\/revisions"}],"predecessor-version":[{"id":4575,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1539\/revisions\/4575"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1539"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1539"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1539"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}