{"id":8274,"date":"2021-11-20T01:17:06","date_gmt":"2021-11-19T19:47:06","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8274"},"modified":"2021-11-20T01:26:01","modified_gmt":"2021-11-19T19:56:01","slug":"find-the-middle-term-in-the-expansion-of-3x-x3over-67","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-middle-term-in-the-expansion-of-3x-x3over-67\/","title":{"rendered":"Find the middle term in the expansion of \\((3x – {x^3\\over 6})^7\\)."},"content":{"rendered":"

Solution :<\/h2>\n

The given expression is \\((3x – {x^3\\over 6})^7\\).<\/p>\n

Here n = 7, which is an odd number.<\/p>\n

By using middle terms in binomial expansion formula<\/a>,<\/p>\n

So, \\(({7 + 1\\over 2}\\)) th and \\(({7 + 1\\over 2} + 1)\\) th i.e.\u00a0 4th and 5th terms are two middle terms.<\/p>\n

Now, \\(T_{4}\\) = \\(T_{3 + 1}\\) = \\(^{7}C_{3}\\) \\((3x)^{7 – 3}\\) \\((-{x^3\\over 6})^{3}\\)<\/p>\n

\\(\\implies\\) \\(T_{4}\\) = \\((-1)^3\\) \\(^{7}C_{3}\\) \\((3x)^{4}\\) \\(({x^3\\over 6})^{3}\\)<\/p>\n

\\(\\implies\\) \\(T_{4}\\) = -35 \\(\\times\\) 81 \\(x^4\\) \\(\\times\\) \\(x^9\\over 216\\) = -\\(105x^{13}\\over 8\\)<\/p>\n

and,\u00a0 \\(T_{5}\\) = \\(T_{5 + 1}\\) = \\(^{7}C_{4}\\) \\((3x)^{7 – 4}\\) \\((-{x^3\\over 6})^{4}\\)<\/p>\n

\\(\\implies\\) \\(T_{5}\\) = \\(^{7}C_{4}\\) \\((3x)^{3}\\) \\(({x^3\\over 6})^{4}\\)<\/p>\n

\\(\\implies\\) \\(T_{5}\\)= 35 \\(\\times\\) 27 \\(x^3\\) \\(\\times\\) \\(x^{12}\\over 1296\\) = \\(35x^{15}\\over 48\\)<\/p>\n

Hence, the middle terms are -\\(105x^{13}\\over 8\\) and \\(35x^{15}\\over 48\\).<\/p>\n

\n

Similar Questions<\/h3>\n

Find the middle term in the expansion of \\(({2\\over 3}x^2 \u2013 {3\\over 2x})^{20}\\).<\/a><\/p>\n

Find the 9th term in the expansion of \\(({x\\over a} \u2013 {3a\\over x^2})^{12}\\).<\/a><\/p>\n

Find the 10th term in the binomial expansion of \\((2x^2 + {1\\over x})^{12}\\).<\/a><\/p>\n

Which is larger \\((1.01)^{1000000}\\) or 10,000?<\/a><\/p>\n

By using binomial theorem, expand \\((1 + x + x^2)^3\\).<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : The given expression is \\((3x – {x^3\\over 6})^7\\). Here n = 7, which is an odd number. By using middle terms in binomial expansion formula, So, \\(({7 + 1\\over 2}\\)) th and \\(({7 + 1\\over 2} + 1)\\) th i.e.\u00a0 4th and 5th terms are two middle terms. Now, \\(T_{4}\\) = \\(T_{3 + …<\/p>\n

Find the middle term in the expansion of \\((3x – {x^3\\over 6})^7\\).<\/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":[177,43],"tags":[],"yoast_head":"\nFind the middle term in the expansion of \\((3x - {x^3\\over 6})^7\\).<\/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-middle-term-in-the-expansion-of-3x-x3over-67\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the middle term in the expansion of \\((3x - {x^3\\over 6})^7\\).\" \/>\n<meta property=\"og:description\" content=\"Solution : The given expression is ((3x – {x^3over 6})^7). 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