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

^{2 n} C_{n}

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\frac{1.3.5 \dots . . (2 n - 1)}{n!} 2^{n}

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2.6 \dots (4 n - 2)

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2.4 \dots \dots \dots 2 n

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Solution

The correct options are
A $^{2 n} C_{n}$
B $\frac{1.3.5 \dots . . (2 n - 1)}{n!} 2^{n}$
Middle term $= \frac{N}{2} + 1$

$= \frac{2 n}{2} + 1$

$= (n + 1) t h t e r m$

Now consider the following

$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$ ...(i)

Where $T$ represents the term.

Here the middle term is the $(n + 1)^{t h}$ term

Hence $r + 1 = n + 1$

$r = n$

substituting in (i), we get

$^{2 n} C_{n} 1^{2 n - n} x^{n}$

$=^{2 n} C_{n} x^{n}$

Hence the coefficient of middle term is $^{2 n} C_{n}$ which can also be written as

$\frac{(1.3.5... (2 n - 1)) 2^{n}}{n!}$

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Similar questions

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

(1 + x)^{2 n}

\frac{1.3.5..... (2 n - 1)}{n!}

2^{n} x^{n}

(1 + x)^{2 n}

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

(1 + x)^{2 n}

Show that the middle term in the expansion of $(1 + x)^{2 n}$ is $\frac{1. 3.5 \dots (2 n - 1)}{n!} . 2^{n} . X^{n}, where n ϵ N .$

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