{"id":6475,"date":"2021-10-16T17:34:15","date_gmt":"2021-10-16T12:04:15","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6475"},"modified":"2021-11-18T22:37:21","modified_gmt":"2021-11-18T17:07:21","slug":"middle-term-in-binomial-expansion","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/middle-term-in-binomial-expansion\/","title":{"rendered":"Middle Term in Binomial Expansion"},"content":{"rendered":"
Here you will learn formula to find middle term in binomial expansion with examples.<\/p>\n
Let’s begin – <\/p>\n
Since the binomial expansion of \\((x + a)^n\\) contains (n + 1) terms. Therefore,<\/p>\n
\n(1) If n is even, then \\({n\\over 2} + 1\\) th term is the middle term.<\/p>\n
(2) If n is odd, then \\({n + 1\\over 2}\\) th and \\({n + 3\\over 2}\\) th terms are the two middle terms.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the middle term in the expansion of \\(({2\\over 3}x^2 – {3\\over 2x})^{20}\\).<\/p>\n
Solution<\/span><\/strong> : Here, n = 20, which is an even number.<\/p>\n
So, \\({20\\over 2} + 1\\)th term i.e. 11th term is the middle term.<\/p>\n
Hence, the middle term = \\(T_{11}\\)<\/p>\n
\\(T_{11}\\) = \\(T_{10 + 1}\\) = \\(^{20}C_{10}\\) \\(({2\\over 3}x^2)^{20 – 10}\\) \\((-{3\\over 2x})^{10}\\)<\/p>\n
= \\(^{20}C_{10} x^{10}\\)<\/p>\n
Example<\/span><\/strong> : Find the middle term in the expansion of \\((3x – {x^3\\over 6})^7\\).<\/p>\n
Solution<\/span><\/strong> : The given expression is \\((3x – {x^3\\over 6})^7\\).<\/p>\n
Here n = 7, which is an odd number.<\/p>\n
So, \\(({7 + 1\\over 2}\\)) th and \\(({7 + 1\\over 2} + 1)\\) th i.e. 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, \\(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\n