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Question: The middle term in the expansion of \((1 + x)^{2n}\) is...

The middle term in the expansion of (1+x)2n(1 + x)^{2n} is

A

1.3.5......(2n−1)n!x2n+1\frac{1.3.5......(2n - 1)}{n!}x^{2n + 1}

B

2.4.6......2nn!x2n+1\frac{2.4.6......2n}{n!}x^{2n + 1}

C

1.3.5......(2n−1)n!xn\frac{1.3.5......(2n - 1)}{n!}x^{n}

D

1.3.5......(2n−1)n!xn.2n\frac{1.3.5......(2n - 1)}{n!}x^{n}.2^{n}

Answer

1.3.5......(2n−1)n!xn.2n\frac{1.3.5......(2n - 1)}{n!}x^{n}.2^{n}

Explanation

Solution

Since 2n is even, so middle term = T2n2+1=Tn+1T_{\frac{2n}{2} + 1} = T_{n + 1}

⇒ Tn+1=2n⥂Cnxn=(2n)!n!.n!xnT_{n + 1} =^{2n} ⥂ C_{n}x^{n} = \frac{(2n)!}{n!.n!}x^{n} = 1.3.5........(2n−1)n!.2nxn\frac{1.3.5........(2n - 1)}{n!}.2^{n}x^{n}.