1. **Problem:** Show that the middle term in the expansion of $ (1+x)^{2n} $ is given by $$ \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{n!} x^n $$ 2. **Formula and Explanation:** The general term in the binomial expansion of $ (1+x)^{2n} $ is $$ T_{k+1} = \binom{2n}{k} x^k $$ where $k=0,1,2,\ldots,2n$. The middle term corresponds to $k=n$ (since there are $2n+1$ terms). 3. **Intermediate Work:** The coefficient of the middle term is $$ \binom{2n}{n} = \frac{(2n)!}{n! \times n!} $$ We want to show that $$ \binom{2n}{n} = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{n!} \times 2^n $$ Note that the product of odd numbers up to $2n-1$ is called the double factorial: $$ (2n-1)!! = 1 \times 3 \times 5 \times \cdots \times (2n-1) $$ Using the identity: $$ (2n)! = 2^n n! (2n-1)!! $$ Divide both sides by $n! n!$: $$ \binom{2n}{n} = \frac{(2n)!}{n! n!} = \frac{2^n n! (2n-1)!!}{n! n!} = \frac{2^n (2n-1)!!}{n!} $$ Therefore, $$ \binom{2n}{n} = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{n!} 2^n $$ 4. **Conclusion:** The middle term is $$ T_{n+1} = \binom{2n}{n} x^n = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{n!} 2^n x^n $$ If the problem states the middle term as $\frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{n!} x^n x^n$, note that $x^n x^n = x^{2n}$, which is the last term, so the correct middle term coefficient includes the factor $2^n$. Hence, the middle term coefficient is $$ \binom{2n}{n} = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{n!} 2^n $$ and the middle term is $$ \binom{2n}{n} x^n = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{n!} 2^n x^n $$ This completes the proof.