text: # The middle term in the expansion of
  
  Question:
  
  The middle term in the expansion of $\left(x-\frac{1}{x}\right)^{18}$ is ______________
  
  Solution:
  
  $\left(x-\frac{1}{x}\right)^{18}$
  
  Since 18 is even
  
  there is only one middle term i.e. $\left(\frac{2 n}{2}+1\right)^{\text {th }}$ term
  
  i.e. (+ 1)th term
  
  i.e. $T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(\frac{-1}{x}\right)^{n}$
  
  $={ }^{2 n} C_{n}(-1)^{n} x^{2 n-n} x^{-n}$
  
  $={ }^{2 n} C_{n}(-1)^{n} x^{0} \quad$ i. e. ${ }^{18} C_{9}(-1)^{18}={ }^{18} C_{9}$
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