An identity satisfied by the middle binomial coefficient

Question

Is there a citable reference for the identity

$$\frac{ \binom{n}{n/2+1}-\binom{n}{n/2+2} }{ \binom{n}{n/2}-\binom{n}{n/2-1} }=\frac{3n}{n+4}$$

for every even natural number $n\geq 1$?

I know how to prove this, and am not asking for a proof, only for a reference, because the proof I found is unpleasantly lengthy.

Answer 1

The following identity holds (see the fourth one HERE): $\binom{n}{k}=\frac{n+1-k}{k}\binom{n}{k-1}$. Hence $$\binom{n}{n/2+2}=\frac{n+1-(n/2+2)}{n/2+2}\binom{n}{n/2+1}=\frac{n-2}{n+4}\binom{n}{n/2+1}$$ and $$\binom{n}{n/2}=\frac{n/2+1}{n+1-(n/2+1)}\binom{n}{n/2+1}=\frac{n+2}{n}\binom{n}{n/2+1}$$ Hence $$\frac{ \binom{n}{n/2+1}-\binom{n}{n/2+2} }{ \binom{n}{n/2}-\binom{n}{n/2-1} }=\frac{ \binom{n}{n/2+1}-\frac{n-2}{n+4}\binom{n}{n/2+1} }{ \frac{n+2}{n}\binom{n}{n/2+1} -\binom{n}{n/2+1} }=\frac{1-\frac{n-2}{n+4}}{\frac{n+2}{n}-1}=\frac{3n}{n+4}.$$

Answer 2

The proof certainly isn't immediate, however it doesn't seem very lengthy (to me, at least)...

$$\frac{ \binom{n}{n/2+1}-\binom{n}{n/2+2} }{ \binom{n}{n/2}-\binom{n}{n/2-1} }=\frac{n(3n+6)}{(n+2)(n+4)}$$

For even $n$ let's substitute $n=2k$:

$$\frac{ \binom{2k}{k+1}-\binom{2k}{k+2} }{ \binom{2k}k-\binom{2k}{k-1} } = \frac{2k(6k+6)}{(2k+2)(2k+4)}$$ Now we expand binomial coefficients with ratios of respective factorials on LHS and reduce RHS a bit: $$\frac{ \frac{(2k)!}{(k+1)!(2k-(k+1))!}-\frac{(2k)!}{(k+2)!(2k-(k+2))!} }{ \frac{(2k)!}{k!(2k-k)!}-\frac{(2k)!}{(k-1)!(2k-(k-1))!} } = \frac{12k(k+1)}{4(k+1)(k+2)}$$ $$\frac{ \frac{(2k)!}{(k+1)!(k-1)!}-\frac{(2k)!}{(k+2)!(k-2)!} }{ \frac{(2k)!}{k!k!}-\frac{(2k)!}{(k-1)!(k+1)!} } = \frac{3k(k+1)}{(k+1)(k+2)}$$

Let's reduce $(2k!)$ from enumerators of all the terms on LHS and $(k+1)$ on RHS:

$$\frac{ \frac 1{(k+1)!(k-1)!}-\frac 1{(k+2)!(k-2)!} }{ \frac 1{k!k!}-\frac 1{(k-1)!(k+1)!} } = \frac{3k}{k+2}$$

Now let's reduce $k!k!$ from denominators:

$$\frac{ \frac k{k+1}-\frac {(k-1)k}{(k+1)(k+2)} }{ \frac 11-\frac k{k+1} } = \frac{3k}{k+2}$$ Now $(k+1)(k+2)$ from denominators: $$\frac{ k(k+2)- (k-1)k }{ (k+1)(k+2) - k(k+2) } = \frac{3k}{k+2}$$ $$\frac{ [(k+2)- (k-1)]k }{ [(k+1)-k](k+2) } = \frac{3k}{k+2}$$ $$\frac{ 3\cdot k }{ 1\cdot(k+2) } = \frac{3k}{k+2}$$ The expression equivalent to the initial one appears an identity.