Correct Answer - Option 3 : 80x and
\(\rm \frac {40}{x}\)
Concept:
General term: General term in the expansion of (x + y)n is given by
\(\rm {T_{\left( {r\; + \;1} \right)}} = \;{\;^n}{C_r} × {x^{n - r}} × {y^r}\)
Middle terms: The middle terms is the expansion of (x + y) n depends upon the value of n.
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If n is even, then total number of terms in the expansion of (x + y) n is n +1. So there is only one middle term i.e. \(\rm \left( {\frac{n}{2} + 1} \right){{\rm{\;}}^{th}}\) term is the middle term.
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If n is odd, then total number of terms in the expansion of (x + y) n is n + 1. So there are two middle terms i.e. \(\rm {\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \(\rm {\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) are two middle terms.
Calculation:
Here, we have to find the middle terms in the expansion of \(\rm \left(2x + \frac 1 x \right)^{5}\)
Here n = 5 (n is odd number)
∴ Middle term = \(\rm {\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \(\rm {\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) = 3rd and 4th
T3 = T (2 + 1) = 5C2 × (2x) (5 - 2) × \(\rm \left(\frac {1}{x}\right)^2\) and T4 = T (3 + 1) = 5C3 × (2x) (5 - 3) × \(\rm \left(\frac {1}{x}\right)^3\)
T3 = 5C2 × (23x) and T4 = 5C3 × 22 × \(\rm \frac 1 x\)
T3 = 80x and T4 = \(\rm \frac {40}{x}\)
Hence the middle term of expansion is 80x and \(\rm \frac {40}{x}\)