Two very difficult induction proofs; having trouble with the inductive step $$\sum_{k=0}^{n-2} \binom{n}{k}\binom{n}{k+1}\frac{n-2k-1}{k+1} = n-2 + \frac{1}{n+1}\binom{2n}{n}$$
$$\sum_{k=0}^{n-2} \binom{n}{k}\binom{n}{k+2}\frac{n-2k-1}{k+1} = -n + \frac{n}{(n+2)(n+1)}\binom{2n}{n}$$
The two are clearly related in some way, so proving one might yield the other, but I'm having a lot of difficulty knowing what to add to both sides to change all those n's to n+1's in the binomial coefficients of the sums. Do any of you have insight?
The induction will be on n, with base case n=2.
I've been using this:
$$\binom{n}{k} = \frac{n+1}{n+1-k}\binom{n+1}{k} $$
as a means of replacing n choose k with some factor of n+1 choose k, though current attempts are fairly circular. Substitute k+1 or k+1 for k to get a change to expand those binomials.
 A: Let me present an algebraic proof of the first equality and a proof by
induction will perhaps appear.
Suppose we seek to verify that
$$\sum_{k=0}^{n-2} {n\choose k} {n\choose k+1}
\frac{n-2k-1}{k+1} = n-2+\frac{1}{n+1}{2n\choose n}.$$
The left has two pieces which are, first, piece $A$,
$$\sum_{k=0}^{n-2} {n\choose k} {n\choose k+1} \frac{n+1}{k+1}
= \sum_{k=0}^{n-2} {n+1\choose k+1} {n\choose k+1}$$
and second, piece $B$,
$$-2\sum_{k=0}^{n-2} {n\choose k} {n\choose k+1}.$$
Piece $A$ is
$$\sum_{k=1}^{n-1} {n+1\choose k} {n\choose k}
= -1 - (n+1) + \sum_{k=0}^{n}  {n\choose k} {n+1\choose k}.$$
Introduce for the sum term
$${n+1\choose k} = {n+1\choose n+1-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+2-k}} (1+z)^{n+1} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+2}} (1+z)^{n+1} 
\sum_{k=0}^{n} {n\choose k} z^k\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+2}} (1+z)^{n+1} 
(1+z)^n \; dz
\\ = {2n+1\choose n+1}.$$
Therefore we have for piece $A$
$$-n-2 + {2n+1\choose n+1}.$$
Piece $B$ is
$$2n - 2\sum_{k=0}^{n-1} {n\choose k} {n\choose k+1}.$$
Introduce for the sum term
$${n\choose k+1} = {n\choose n-k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k}} (1+z)^{n} \; dz.$$
This is zero when  $k=n$ so we may include this term  in the sum which
yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n}} (1+z)^{n}
\sum_{k=0}^n {n\choose k} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n}} (1+z)^{n}
(1+z)^n
\; dz
\\ = {2n\choose n-1}.$$
This yields for piece $B$
$$2n-2{2n\choose n+1}.$$
Collecting the two pieces we have
$$n-2 + {2n+1\choose n+1} - 2{2n\choose n+1}
\\ = n-2 + \frac{2n+1}{n+1} {2n\choose n}
- 2 \frac{n}{n+1} {2n\choose n}
\\ = n-2 + \frac{1}{n+1} {2n\choose n}.$$
