{"id":6871,"date":"2021-10-21T21:28:58","date_gmt":"2021-10-21T15:58:58","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6871"},"modified":"2021-10-25T02:26:05","modified_gmt":"2021-10-24T20:56:05","slug":"if-sum_r1n%e2%80%8e-t_r-nover-8-n-1n-2n-3-then-find-sum_r1n%e2%80%8e-1over-t_r","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/if-sum_r1n%e2%80%8e-t_r-nover-8-n-1n-2n-3-then-find-sum_r1n%e2%80%8e-1over-t_r\/","title":{"rendered":"If \\({\\sum}_{r=1}^{n\u200e} T_r\\) = \\(n\\over 8\\) (n + 1)(n + 2)(n + 3), then find \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\)"},"content":{"rendered":"

Solution :<\/h2>\n

\\(\\because\\) \\(T_n\\) = \\(S_n – S_{n-1}\\)<\/p>\n

= \\({\\sum}_{r=1}^{n\u200e} T_r\\) – \\({\\sum}_{r=1}^{n\u200e – 1} T_r\\)<\/p>\n

= \\(n(n+1)(n+2)(n+3)\\over 8\\) – \\((n-1)(n)(n+1)(n+2)\\over 8\\)<\/p>\n

= \\(n(n+1)(n+2)\\over 8\\)[(n+3) – (n-1)] = \\(n(n+1)(n+2)\\over 8\\)(4)<\/p>\n

\\(T_n\\) = \\(n(n+1)(n+2)\\over 2\\)<\/p>\n

\\(\\implies\\) \\(1\\over T_n\\) = \\(2\\over n(n+1)(n+2)\\) = \\((n+2)-n\\over n(n+1)(n+2)\\)<\/p>\n

= \\(1\\over n(n+1)\\) – \\(1\\over (n+1)(n+2)\\)<\/p>\n

Let \\(V_n\\) = \\(1\\over n(n+1)\\)<\/p>\n

\\(\\therefore\\) \\(1\\over T_n\\) = \\(V_n\\) – \\(V_{n+1}\\)<\/p>\n

Putting n = 1, 2, 3, …… n<\/p>\n

\\(\\implies\\) \\(1\\over T_1\\) + \\(1\\over T_2\\) + \\(1\\over T_3\\) + ….. + \\(1\\over T_n\\) = \\(V_1\\) – \\(V_{n+1}\\)<\/p>\n

= \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\) = \\(n^2 + 3n\\over 2(n+1)(n+2)\\)<\/p>\n


\n

Similar Questions<\/h3>\n

Let \\(a_n\\) be the nth term of an AP. If \\(\\sum_{r=1}^{100}\\) \\(a_{2r}\\) = \\(\\alpha\\) and \\(\\sum_{r=1}^{100}\\) \\(a_{2r-1}\\) = \\(\\beta\\), then the common difference of the AP is<\/a><\/p>\n

Prove that the sum of first n natural numbers is \\(n(n+1)\\over 2\\)<\/a><\/p>\n

If \\((10)^9\\) + \\(2(11)^1(10)^8\\) + \\(3(11)^2(10)^7\\) + \u2026\u2026 + \\(10(11)^9\\) = \\(K(10)^9\\), then k is equal to<\/a><\/p>\n

If x, y and z are in AP and \\(tan^{-1}x\\), \\(tan^{-1}y\\) and \\(tan^{-1}z\\) are also in AP, then<\/a><\/p>\n

Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + \u2026\u2026<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : \\(\\because\\) \\(T_n\\) = \\(S_n – S_{n-1}\\) = \\({\\sum}_{r=1}^{n\u200e} T_r\\) – \\({\\sum}_{r=1}^{n\u200e – 1} T_r\\) = \\(n(n+1)(n+2)(n+3)\\over 8\\) – \\((n-1)(n)(n+1)(n+2)\\over 8\\) = \\(n(n+1)(n+2)\\over 8\\)[(n+3) – (n-1)] = \\(n(n+1)(n+2)\\over 8\\)(4) \\(T_n\\) = \\(n(n+1)(n+2)\\over 2\\) \\(\\implies\\) \\(1\\over T_n\\) = \\(2\\over n(n+1)(n+2)\\) = \\((n+2)-n\\over n(n+1)(n+2)\\) = \\(1\\over n(n+1)\\) – \\(1\\over (n+1)(n+2)\\) Let \\(V_n\\) = \\(1\\over n(n+1)\\) \\(\\therefore\\) …<\/p>\n

If \\({\\sum}_{r=1}^{n\u200e} T_r\\) = \\(n\\over 8\\) (n + 1)(n + 2)(n + 3), then find \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\)<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,57],"tags":[],"yoast_head":"\nIf \\({\\sum}_{r=1}^{n\u200e} T_r\\) = \\(n\\over 8\\) (n + 1)(n + 2)(n + 3), then find \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\)<\/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:\/\/mathemerize.com\/if-sum_r1n\u200e-t_r-nover-8-n-1n-2n-3-then-find-sum_r1n\u200e-1over-t_r\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"If \\({\\sum}_{r=1}^{n\u200e} T_r\\) = \\(n\\over 8\\) (n + 1)(n + 2)(n + 3), then find \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\)\" \/>\n<meta property=\"og:description\" content=\"Solution : (because) (T_n) = (S_n – S_{n-1}) = ({sum}_{r=1}^{n\u200e} T_r) – ({sum}_{r=1}^{n\u200e – 1} T_r) = (n(n+1)(n+2)(n+3)over 8) – ((n-1)(n)(n+1)(n+2)over 8) = (n(n+1)(n+2)over 8)[(n+3) – (n-1)] = (n(n+1)(n+2)over 8)(4) (T_n) = (n(n+1)(n+2)over 2) (implies) (1over T_n) = (2over n(n+1)(n+2)) = ((n+2)-nover n(n+1)(n+2)) = (1over n(n+1)) – (1over (n+1)(n+2)) Let (V_n) = (1over n(n+1)) (therefore) … If ({sum}_{r=1}^{n\u200e} T_r) = (nover 8) (n + 1)(n + 2)(n + 3), then find ({sum}_{r=1}^{n\u200e} )(1over T_r) Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathemerize.com\/if-sum_r1n\u200e-t_r-nover-8-n-1n-2n-3-then-find-sum_r1n\u200e-1over-t_r\/\" \/>\n<meta property=\"og:site_name\" content=\"Mathemerize\" \/>\n<meta property=\"article:published_time\" content=\"2021-10-21T15:58:58+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-10-24T20:56:05+00:00\" \/>\n<meta name=\"author\" content=\"mathemerize\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"mathemerize\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<script type=\"application\/ld+json\" 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