If r th term is the middle term in the expansion of \(\Big(\text{x}^{2}-\frac{1}{2\text{x}}\Big)^{20},\)…

If rth term is the middle term in the expansion of \(\Big(\text{x}^{2}-\frac{1}{2\text{x}}\Big)^{20},\) then (r + 3)th term is:

  1. \({^\text{20}}\text{C}_{\text{14}}\ \Big(\frac{\text{x}}{2^{14}}\Big)\)

  2. \({^\text{20}}\text{C}_{\text{12}}\ \text{x}^{2}\ 2^{-12}\)

  3. \(-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}\)

  4. None of these.

Solution

Solution:

Here, n is even,

So, The middle term in the given expansion is \(\Big(\frac{20}{2}+1\Big)^{\text{th}}=11^{\text{th}}\)

Therefore, (r + 3)th term is the 14th term

\(\text{T}_{14}={^\text{20}}\text{C}_{\text{13}}(\text{x}^{2})^{20-13}\ \Big(\frac{-1}{2\text{x}}\Big)\)

\(=(-1)^{13}\ {^\text{20}}\text{C}_{\text{13}}\ \frac{\text{x}^{14-3}}{2^{13}}\)

\(=-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}\)

Asked in: School

Practice more Binomial Theorem questions on Aicharya