What is the coefficient of the middle term in the expansion of (1 + 4x + 4x²)⁵?

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.

(1−x)n=∑k=0n(nk)1n−k(−x)k(1−x)n=∑k=0n(nk)1n−k(−x)k(1−x)n=∑k=0n(nk)1n−k(−x)

Calculation:

Given:

(1 + 4x + 4x2)5

⇒ [(1 + 2x)²]⁵

⇒ (1+ 2x)¹⁰ 

Here n = 10 (n is even number)

∴ Middle term = \(\left( {\frac{n}{2} + 1} \right) = \left( {\frac{10}{2} + 1} \right) \) = 6th term 

Middle Term, T₆ =  T_{5 + 1} = ¹⁰C₅ (1)⁵ (2x)⁵

⇒ \(\frac {10!}{5!5!}\)× 32x⁵

⇒ 8064 x⁵

∴ The coefficient of the middle term in the expansion of (1 + 4x + 4x2)5  is 8064.