Question8. (a) Verify that for all $n \geq 1$, $$2 \cdot 6 \cdot 10 \cdot 14 \cdots(4 n-2)=\frac{(2 n)!}{n!}$$ (b) Use part (a) to obtain the inequality $2^{n}(n!)^{2} \leq(2 n)$ ! for all $n \geq 1$.

Studdy Solution
Now, use the verified equation to derive the inequality.
Given:
$$2 \cdot 6 \cdot 10 \cdot 14 \cdots (4n - 2) = \frac{(2n)!}{n!}$$
We can rewrite the left-hand side as:
$$2^n \cdot n!$$
Thus, we have:
$$2^n \cdot n! = \frac{(2n)!}{n!}$$
Multiplying both sides by $n!$, we get:
$$2^n \cdot (n!)^2 = (2n)!$$
This implies:
$$2^n \cdot (n!)^2 \leq (2n)!$$
The inequality is:
$$2^n (n!)^2 \leq (2n)!$$