The middle term in the expansion of \(\Big(\frac{2\text{x}}{3}=\frac{3}{2\text{x}^{2}}\Big)^{2\text{n}}\) is:

The middle term in the expansion of \(\Big(\frac{2\text{x}}{3}=\frac{3}{2\text{x}^{2}}\Big)^{2\text{n}}\) is:

  1. None of these.

  2. \({^\text{2n}}\text{C}_{\text{n}}\)

  3. \({^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}\)

  4. \((-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}\)

Solution

Solution:

Here, n is even,

Middle term in the given expansion \(=\Big(\frac{2\text{n}}{2}+1\Big)^{\text{th}}=(\text{n}+1)\)

\(={^\text{2n}}\text{C}_{\text{n}}\Big(\frac{2\text{x}}{3}\Big)^{2\text{n}-\text{n}}\ \Big(\frac{-3}{2\text{x}^{2}}\Big)^{\text{n}}\)

\((-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}\)

Asked in: School

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