Show that the middle term in the expansion of $(1+x)^{2n}$ is $\dfrac{1.3.5.\cdots.(2n-1)}{n!} 2^n x^n$, where $n$ is a positive integer.

Question

Accepted Answer

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> [!TLDR]
> Split $(2n)!$ into its odd and even factors, then simplify $\binom{2n}{n}$.
>
> $(1+x)^{2n}$ has $2n+1$ terms, so its middle term is the $(n+1)$th term, $T_{n+1}$.
>
> $T_{n+1} = {}^{2n}C_n\, x^n = \dfrac{(2n)!}{n!\,n!}x^n$
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> Now split $(2n)! = 1\cdot2\cdot3\cdots(2n)$ into its odd-position and even-position factors:
> $(2n)! = \underbrace{[1\cdot3\cdot5\cdots(2n-1)]}_{	ext{odd factors}}	imes\underbrace{[2\cdot4\cdot6\cdots(2n)]}_{	ext{even factors}}$
>
> The even factors can be written as $2\cdot4\cdots(2n) = 2^n(1\cdot2\cdot3\cdots n) = 2^n\,n!$
>
> So $(2n)! = [1\cdot3\cdot5\cdots(2n-1)]\cdot2^n\cdot n!$
>
> Therefore: …

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Q.Show that the middle term in the expansion of $(1+x)^{2n}$ is $\dfrac{1.3.5.\cdots.(2n-1)}{n!} 2^n x^n$ , where $n$ is a positive integer.

Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

The Pattern You Already Know

The Precise Statement

Where Do Those Coefficients Come From?

The General Term

What If It's Not Just $x$ and $y$ ? …

Q.Show that the middle term in the expansion of (1+x)2n(1+x)^{2n}(1+x)2n is 1.3.5.⋯ .(2n−1)n!2nxn\dfrac{1.3.5.\cdots.(2n-1)}{n!} 2^n x^nn!1.3.5.⋯.(2n−1)​2nxn, where nnn is a positive integer.

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Q.Show that the middle term in the expansion of $(1+x)^{2n}$ is $\dfrac{1.3.5.\cdots.(2n-1)}{n!} 2^n x^n$ , where $n$ is a positive integer.

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