Show that $3\cdot \frac{n!}{(n-2)!} + 210 = \frac{(2n)!}{(2n-2))!}$ for a specific n.

Question

Show that:

$$3\cdot \frac{n!}{(n-2)!} + 210 = \frac{(2n)!}{(2n-2)!}$$ has the solution n = 14 and also show that this is the only solution.

I don´t know where to start, the only reason I know that n = 14 is a solution is that I used the trial.

Thanks!

/Alex

Answer 1

$$3\cdot \frac{n!}{(n-2)!} + 210 = \frac{(2n)!}{(2n-2)!}$$

Note that $$\frac{n!}{(n-2)!}=n(n-1)$$

and $$ \frac{(2n)!}{(2n-2)!}= 2n(2n-1) $$

Upon substitution and simplification, we get

$$n^2+n-210=0$$ Thus $n=14$ or $n=-15.$

The only positive solution is $n=14.$

Answer 2

Write $\frac{n!}{(n-2)!}=n(n-1)$ and similarly $\frac{(2n)!}{(2n-2)!}=2n(2n-1)$ to get a quadratic equation in $n$.