How to show that these products are equal?

Question

I need your help. I'm trying to show that these products are equal:

$$\prod_{k=1}^{n}(4k-2)=\prod_{k=1}^{n}(n+k)$$

Thank you in advance !

PS: I need two different ways to solve the problem...

Answer 1

$$\prod_{k=1}^n2(2k-1)=2^n\prod_{k=1}^n(2k-1)=\frac{\prod_{k=1}^n(2k-1)\cdot\prod_{k=1}^n(2k)}{n!}=\frac{(2n)!}{n!}$$

$$\prod_{k=1}^n(n+k)=\frac{\prod_{k=1}^n k\cdot\prod_{k=1}^n(n+k)}{n!}=?$$

Answer 2

You said you needed 2 different ways to solve the problem.

One is already given by lab. The other can be good-old induction. Here is the induction step: $$ \prod_{k=1}^{n+1}(4k-2)=\prod_{k=1}^{n}(4k-2)\times(4n+4-2)=\prod_{k=1}^n(n+k)\times (4n+2)\\ =\prod_{k=0}^{n-1}(n+1+k)\times(4n+2). $$ Now note that $4n+2=\frac{(n+1+n)(n+1+n+1)}{n+1+0}$ so the last expression above equals $\prod_{k=1}^{n+1}(n+1+k)$.

Answer 3

Let $f(n)=\prod_{k=1}^{n}(4k-2)$ and let $g(n)=\prod_{k=1}^{n}(n+k)$.

Then $f$ can be expressed recursively as $f(1)=2$ and $f(n+1)=(4n+2)\cdot f(n)$.

And $g$ can be expressed recursively as $g(1)=2$ and $g(n+1)=\frac{g(n)}{n+1}\cdot(2n+1)\cdot(2n+2)=g(n)\cdot(4n+2)$.

So $f$ and $g$ satisfy the same recursion with the same initial condition, and hence are equal.