{"id":8207,"date":"2021-11-17T15:20:50","date_gmt":"2021-11-17T09:50:50","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8207"},"modified":"2021-11-17T15:35:43","modified_gmt":"2021-11-17T10:05:43","slug":"find-the-sum-of-n-terms-of-the-series-3-7-14-24-37","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-sum-of-n-terms-of-the-series-3-7-14-24-37\/","title":{"rendered":"Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + ……."},"content":{"rendered":"

Solution :<\/h2>\n

By using method of differences<\/a>,<\/p>\n

The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. Clearly,these differences are in AP.<\/p>\n

Let \\(T_n\\) be the nth term and \\(S_n\\) denote the sum to n terms of the given series<\/p>\n

Then, \\(S_n\\) = 3 + 7 + 14 + 24 + 37 + ….. + \\(T_{n-1}\\) + \\(T_n\\) ……(i)<\/p>\n

Also, \\(S_n\\) =\u00a0 \u00a0 \u00a0 \u00a0 \u00a03 + 7 + 14 + 24 + 37 + ….. + \\(T_{n-1}\\) + \\(T_n\\) ……(ii)<\/p>\n

Subtracting (ii) from (i), we get<\/p>\n

0 = 3 + [4 + 7 + 10 + 13 + …… + \\(T_{n-1}\\) + \\(T_n\\)] – \\(T_n\\)<\/p>\n

\\(\\implies\\) \\(T_n\\) = 3 + \\((n-1)\\over 2\\){2*4+(n-1-1)*3} = \\(6 + (n-1)(3n+2)\\over 4\\)<\/p>\n

\\(\\implies\\) \\(T_n\\) = \\(1\\over 2\\) (\\(3n^2 – n + 4\\))<\/p>\n

\\(\\therefore\\) \\(S_n\\) = \\(\\sum_{k=1}^{n}\\) \\(T_k\\) = \\(\\sum_{k=1}^{n}\\) \\(1\\over 2\\) (\\(3n^2 – n + 4\\)) = \\(1\\over 2\\)[\\(\\sum_{k=1}^{n}\\)\\(3k^2\\) – \\(\\sum_{k=1}^{n}\\) k + 4n]<\/p>\n

\\(\\implies\\) \\(S_n\\) = \\(1\\over 2\\)[3\\(n(n+1)(2n+1)\\over 6\\)\u00a0 – \\(n(n+1)\\over 2\\) + 4n] = \\(n\\over 2\\) (\\(n^2 + n + 4\\))<\/p>\n

\n

Similar Questions<\/h3>\n

Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + \u2026\u2026<\/a><\/p>\n

Find the sum to n terms of the series : 3 + 15 + 35 + 63 + \u2026..<\/a><\/p>\n

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is<\/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

Three positive integers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then the common ratio of GP is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : By using method of differences, The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. Clearly,these differences are in AP. Let \\(T_n\\) be the nth term and \\(S_n\\) denote the sum to n terms of the given series Then, \\(S_n\\) …<\/p>\n

Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + …….<\/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":"","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":"\nFind the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + .......<\/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\/find-the-sum-of-n-terms-of-the-series-3-7-14-24-37\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + .......\" \/>\n<meta property=\"og:description\" content=\"Solution : By using method of differences, The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. 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