What is the coefficient of the middle term in the binomial expansion of (2 + 3x) ⁴?

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 terms is the expansion of (x + y) ⁿ depends 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 (2 + 3x) ⁴

Here n = 4 (n is even number)

∴ Middle term = \(\left( {\frac{n}{2} + 1} \right) = \left( {\frac{4}{2} + 1} \right) = 3rd\;term\)

T₃ = T _{(2 + 1)} = ⁴C₂ × (2) ^{(4 - 2)} × (3x) ²

T₃ = 6 × 4 × 9x² = 216 x²