1. ## factorials

I am trying to show how to solve this using mathematical induction.
I don't think I quite understand what to do with the factorials:

Prove that 2x6x10...x(4n-2) = (2n)!/n!

2. Hello, Nichelle14!

Using mathematical induction, prove that: . $2 \cdot 6 \cdot 10\,\cdots\,(4n-2) \;= \;\frac{(2n)!}{n!}$
Verify $S(1):\;\;2 \:=\:\frac{(2\cdot1)!}{1!} \:= \:2$ . . . true!

Assume $S(k):\;\;2\cdot6\cdot10\,\cdots\,(4k-2)\;= \;\frac{(2k)!}{k!}$

Multiply both sides by $4(k+1) - 2 \:= \:4k + 2$

. . $2\cdot6\cdot10\,\hdots\,(4k-2)\cdot(4[k+1] - 2) \;= \;\frac{(2k)!}{k!}\cdot(4k + 2)$

The left side is the left side of $S(k+1).$

We must show that the right side is: . $\frac{(2k + 2)!}{(k+1)!}$ ... the right side of $S(k+1).$

We have: . $\frac{(2k)!\cdot2(2k+1)}{k!}$

Multiply top and bottom by $k + 1:\;\;\frac{(2k)!\cdot2(2k+1)}{k!}\cdot\frac{k + 1}{k + 1}$

. . and we have: . $\frac{(2k)!\cdot(2k+1)\cdot2(k+1)}{k!(k+1)} \;=$ $\frac{(2k)!(2k+1)(2k+2)}{(k+1)!} \;= \;\frac{(2k+2)!}{(k+1)!}$

Therefore, we have proved $S(k+1):$

. . $2\cdot6\cdot10\,\cdots\,(4k-2)\cdot(4k + 2) \;= \;\frac{(2k+2)!}{(k+1)!}$ . . . ta-DAA!

3. thanks. I was doing it right, but got stumped until you showed to mulitply by (k+1)