{"id":8208,"date":"2021-11-17T15:19:24","date_gmt":"2021-11-17T09:49:24","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8208"},"modified":"2021-11-17T15:35:32","modified_gmt":"2021-11-17T10:05:32","slug":"find-the-sum-to-n-terms-of-the-series-3-15-35-63","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-sum-to-n-terms-of-the-series-3-15-35-63\/","title":{"rendered":"Find the sum to n terms of the series : 3 + 15 + 35 + 63 + ….."},"content":{"rendered":"

Solution :<\/h2>\n

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

The difference between the successive terms are 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, …. 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 + 15 + 35 + 63 + ….. + \\(T_{n-1}\\) + \\(T_n\\) ……(i)<\/p>\n

Also, \\(S_n\\) =\u00a0 \u00a0 \u00a0 \u00a0 \u00a03 + 15 + 35 + ….. + \\(T_{n-1}\\) + \\(T_n\\) ……(ii)<\/p>\n

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

0 = 3 + [12 + 20 + 28 + …… + \\(T_{n-1}\\) + \\(T_n\\)] – \\(T_n\\)<\/p>\n

\\(\\implies\\) \\(T_n\\) = 3 + \\((n-1)\\over 2\\){2*12+(n-1-1)*8} = 3 + (n-1){12+4n-8}<\/p>\n

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

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

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

\n

Similar Questions<\/h3>\n

Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + \u2026\u2026.<\/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

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 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, …. 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 to n terms of the series : 3 + 15 + 35 + 63 + …..<\/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 to n terms of the series : 3 + 15 + 35 + 63 + .....<\/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-to-n-terms-of-the-series-3-15-35-63\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the sum to n terms of the series : 3 + 15 + 35 + 63 + .....\" \/>\n<meta property=\"og:description\" content=\"Solution : By using method of differences, The difference between the successive terms are 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, …. 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