1. **Problem:** Show that the middle term in the expansion of $ (1 + x)^{2n} $ is given by $$ \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1) \times 2^n \times x^n}{n!} $$ 2. **Formula and Explanation:** The general term (\(r+1\)th term) in the expansion of $ (1 + x)^m $ is $$ T_{r+1} = \binom{m}{r} x^r $$ For $ m = 2n $, the middle term corresponds to $ r = n $ (since total terms are $ 2n + 1 $). 3. **Calculate the middle term:** $$ T_{n+1} = \binom{2n}{n} x^n $$ 4. **Express $ \binom{2n}{n} $ in terms of double factorials:** Recall that $$ \binom{2n}{n} = \frac{(2n)!}{(n!)^2} $$ Also, the double factorial for odd numbers is $$ 1 \times 3 \times 5 \times \cdots \times (2n - 1) = \frac{(2n)!}{2^n n!} $$ 5. **Substitute this into the expression:** $$ \binom{2n}{n} = \frac{(2n)!}{(n!)^2} = \frac{2^n n! \times (1 \times 3 \times 5 \times \cdots \times (2n - 1))}{(n!)^2} = \frac{2^n (1 \times 3 \times 5 \times \cdots \times (2n - 1))}{n!} $$ 6. **Therefore, the middle term is:** $$ T_{n+1} = \binom{2n}{n} x^n = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1) \times 2^n \times x^n}{n!} $$ This completes the proof. --- **Final answer:** The middle term in the expansion of $ (1 + x)^{2n} $ is $$ \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1) \times 2^n \times x^n}{n!} $$