{"id":8267,"date":"2021-11-20T01:01:55","date_gmt":"2021-11-19T19:31:55","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8267"},"modified":"2021-11-20T01:31:00","modified_gmt":"2021-11-19T20:01:00","slug":"find-the-10th-term-in-the-binomial-expansion-of-2x2-1over-x12","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-10th-term-in-the-binomial-expansion-of-2x2-1over-x12\/","title":{"rendered":"Find the 10th term in the binomial expansion of \\((2x^2 + {1\\over x})^{12}\\)."},"content":{"rendered":"

Solution :<\/h2>\n

We know that the (r + 1)th term or general term in the expansion<\/a> of \\((x + a)^n\\) is given by<\/p>\n

\\(T_{r + 1}\\) = \\(^{n}C_r x^{n – r} a^r\\)<\/p>\n

In the expansion of \\((2x^2 + {1\\over x})^{12}\\), we have<\/p>\n

\\(T_{10}\\) = \\(T_{9 + 1}\\) = \\(^{12}C_9 (2x^2)^{12 – 9} ({1\\over x})^9\\)<\/p>\n

\\(\\implies\\) \\(T_{10}\\) = \\(^{12}C_9 (2x^2)^3 {1\\over x^9}\\)<\/p>\n

\\(\\implies\\) \\(T_{10}\\) = \\(^{12}C_9\\) \\(8\\over x^3\\) = \\(1760\\over x^3\\)<\/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 middle term in the expansion of \\((3x \u2013 {x^3\\over 6})^7\\).<\/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 : We know that the (r + 1)th term or general term in the expansion of \\((x + a)^n\\) is given by \\(T_{r + 1}\\) = \\(^{n}C_r x^{n – r} a^r\\) In the expansion of \\((2x^2 + {1\\over x})^{12}\\), we have \\(T_{10}\\) = \\(T_{9 + 1}\\) = \\(^{12}C_9 (2x^2)^{12 – 9} ({1\\over x})^9\\) \\(\\implies\\) …<\/p>\n

Find the 10th term in the binomial expansion of \\((2x^2 + {1\\over x})^{12}\\).<\/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 10th term in the binomial expansion of \\((2x^2 + {1\\over x})^{12}\\).<\/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-10th-term-in-the-binomial-expansion-of-2x2-1over-x12\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the 10th term in the binomial expansion of \\((2x^2 + {1\\over x})^{12}\\).\" \/>\n<meta property=\"og:description\" content=\"Solution : We know that the (r + 1)th term or general term in the expansion of ((x + a)^n) is given by (T_{r + 1}) = (^{n}C_r x^{n – r} a^r) In the expansion of ((2x^2 + {1over x})^{12}), we have (T_{10}) = (T_{9 + 1}) = (^{12}C_9 (2x^2)^{12 – 9} ({1over x})^9) (implies) … Find the 10th term in the binomial expansion of ((2x^2 + {1over x})^{12}). 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