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.  4th and 5th terms are two middle terms. Now, \(T_{4}\) = \(T_{3 +

Solution : Here, n = 20, which is an even number. So, \({20\over 2} + 1\)th term i.e. 11th term is the middle term. Hence, the middle term = \(T_{11}\) \(T_{11}\) = \(T_{10 + 1}\) = \(^{20}C_{10}\) \(({2\over 3}x^2)^{20 – 10}\) \((-{3\over 2x})^{10}\) = \(^{20}C_{10} x^{10}\) Similar Questions Find the middle term in the expansion

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 \(({x\over a} – {3a\over x^2})^{12}\), we have \(T_{9}\) = \(T_{8 + 1}\) = \(^{12}C_8 ({x\over a})^{12 – 8} ({-3a\over

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 \(T_{10}\) = \(T_{9 + 1}\) = \(^{12}C_9 (2x^2)^{12 – 9} ({1\over x})^9\) \(\implies\)

Solution : We have, \((1.01)^{1000000}\)  – 10000 = \((1 + 0.01)^{1000000}\) – 10000 By using binomial theorem, = \(^{1000000}C_0\) + \(^{1000000}C_1 (0.01)\)  + \(^{1000000}C_2 (0.01)^2\)  + …… + \(^{1000000}C_{1000000} (0.01)^{1000000}\) – 10000 = (1 + 1000000(0.01) + other positive terms) – 10000 = (1 + 10000 + other positive terms) – 10000 = 1 +

Solution : Let y = x + \(x^2\). Then, \((1 + x + x^2)^3\) = \((1 + y)^3\) = \(^3C_0\) + \(^3C_1 y\) + \(^3C_2 y^2\) + \(^3C_3 y^3\) = \(1 + 3y + 3y^2 + y^3\) = 1 + 3\((x + x^2)\) + 3\((x + x^2)^2\) + \((x + x^2)^3\) = \(x^6 + 3x^5