Question8. (a) Verify that for all $n \geq 1$, $$2 \cdot 6 \cdot 10 \cdot 14 \cdots \cdots(4 n-2)=\frac{(2 n)!}{n!}$$

Studdy Solution
Prove for $n = k + 1$: The product for $n = k+1$ is: $$2 \cdot 6 \cdot 10 \cdot \ldots \cdot (4k-2) \cdot (4(k+1)-2)$$
Using the induction hypothesis: $$= \frac{(2k)!}{k!} \cdot (4k + 2)$$
We need to show: $$\frac{(2(k+1))!}{(k+1)!} = \frac{(2k)!}{k!} \cdot (4k + 2)$$
Expanding the left-hand side: $$\frac{(2k+2)!}{(k+1)!} = \frac{(2k+2)(2k+1)(2k)!}{(k+1)k!}$$
Simplifying: $$= \frac{(2k+2)(2k+1)}{k+1} \cdot \frac{(2k)!}{k!}$$
Notice: $$\frac{(2k+2)(2k+1)}{k+1} = 4k + 2$$
Thus: $$\frac{(2k+2)!}{(k+1)!} = \frac{(2k)!}{k!} \cdot (4k + 2)$$
This completes the induction step, proving the formula for $n = k+1$.
The formula is verified for all $n \geq 1$.