Show that for all $n \geq 2, \binom{2n-1}{n} - \binom{2n-1}{n-2} = \frac{1}{n+1}\binom{2n}{n}$

Question

Show that for all $n \geq 2, \binom{2n-1}{n} - \binom{2n-1}{n-2} = \frac{1}{n+1}\binom{2n}{n}$. I'm not sure how to get started on this problem or what theorems or identities I need to use.

Answer 1

Consider these facts:

1. $$\binom{n}{k} = \binom{n}{n-k}$$
2. $$\binom{n-1}{k-1} + \binom{n-1}{k} = \binom{n}{k}$$

From the first fact, we have:

$$\binom{2n-1}{n} = \binom{2n-1}{n-1}$$ and $$\binom{2n-1}{n-2} = \binom{2n-1}{n+1}$$ From the second fact, we have: $$\binom{2n-1}{n+1} = \binom{2n}{n+1} - \binom{2n-1}{n}$$ combining these findings, the LHS becomes: $$\binom{2n-1}{n-1} - [\binom{2n}{n+1} - \binom{2n-1}{n}]$$ Again, after doing the subtraction and by the second fact, the LHS is reduced to:

$$\binom{2n}{n} - \binom{2n}{n+1}$$

Now, using the definition of $\binom{n}{k}$, we have:

$$\frac{2n!}{n!\times n!} - \frac{2n!}{(n+1)!\times (n-1)!}$$ implying: $$\frac{(n+1)2n! - (n)2n!}{(n+1)!\times n!}$$ Which is nothing but: $$\frac{2n!}{(n+1)! n!}$$ Now, extracting the $\frac{1}{n+1}$, we get: $$\frac{1}{n+1} \times \frac{2n!}{n! n!} = \frac{1}{n+1} \binom{2n}{n}$$ which is the result on the RHS.

Answer 2

As suggested by D.R., directly by the definition we have

$$\binom{2n-1}{n} - \binom{2n-1}{n-2} = \frac{1}{n+1}\binom{2n}{n}$$

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

$$\frac{(n+1)n!n!}{(2n-1)!}\left(\frac{(2n-1)!}{n!(n-1)!} - \frac{(2n-1)!}{(n-2)!(n+1)!} \right)= \frac{(n+1)n!n!}{(2n-1)!}\frac{1}{n+1}\frac{2n(2n-1)!}{n!n!}$$

$$n(n+1) - n(n-1) = 2n$$