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Find the middle term of the expansion of \(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^8\)
Concept:General term: General term in the expansion of (x + y)n is given by\({T_{\left( {r\; + \;1} \right)}} = \;{\;^n}{C_r} \times {x^{n - r}} \times {y^r}\)Middle terms: The middle term is the expansion of (x + y) n depending upon the value of n.If n is even, then the total number of terms in the expansion of (x + y) n is n +1. So there is only one middle term i.e. \(\left( {\frac{n}{2} + 1} \right){{\rm{\;}}^{th}}\) term is the middle term.\({T_{\left( {\frac{n}{2}\; + \;1} \right)}} = \;{\;^n}{C_{\frac{n}{2}}} \times {x^{\frac{n}{2}}} \times {y^{\frac{n}{2}}}\)If n is odd, then the total number of terms in the expansion of (x + y) n is n + 1. So there are two middle terms i.e. \({\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \({\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) are two middle terms. Calculation:Here, we have to find the coefficient of the middle term in the binomial expansion of \(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^8\)Here n = 8 (n is even number)? Middle term = \(\left( {\frac{n}{2} + 1} \right) = \left( {\frac{8}{2} + 1} \right) = 5^{th}\ term\)Hence, middle terms or 5th term will be,T5 = \(C^8_4\times (\frac{x}{y})^4\times (\frac{y}{x})^4=\ ^8C_4\)
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