Question(a) Verify that for all $n \geq 1$, $$2 \cdot 6 \cdot 10 \cdot 14 \cdots \cdots(4 n-2)=\frac{(2 n)!}{n!}$$ (b) Use part (a) to obtain the inequality $2^{n}(n!)^{2} \leq(2 n)$ ! for all $n \geq 1$.

Studdy Solution
Using the verified equality, we have: $$2 \cdot 6 \cdot 10 \cdot \cdots \cdot (4n - 2) = \frac{(2n)!}{n!}$$
This implies: $$2^n \cdot n! = \frac{(2n)!}{n!}$$
Rearranging gives: $$2^n \cdot (n!)^2 \leq (2n)!$$
The inequality derived is:
$$2^n (n!)^2 \leq (2n)!$$