It is not so easy as I had hoped. But the initial work can be done without induction.

$Given:\ k,\ n,\ r \in \mathbb Z\ and\ 1 \le k \le r \le n - 1.$

$Define\ \displaystyle x(k,\ n,\ r) = \sum_{m=k+1}^n \left \{ \dfrac{m - k}{n - r} * \dbinom{m}{k} * \dbinom{n - m}{r - k} \div \dbinom{n + 1}{r + 1} \right \}.$

$Prove:\ x(k,\ n,\ r) = \dfrac{k + 1}{r + 1}.$

We need a lemma, which is step 1.

$Step\ 1:\ \alpha \in \mathbb Z\ and\ \alpha \ge 2 \implies$

$\displaystyle \left ( \sum_{i= 2}^{\alpha}(i^2 - i) \right ) = \left (\sum_{i= 2}^{\alpha}i^2 \right ) - \left ( \sum_{i= 2}^{\alpha}i \right ) =\left (-\ 1^2 + \sum_{i= 1}^{\alpha}i^2 \right ) - \left (-\ 1 + \sum_{i= 1}^{\alpha}i \right ) =$

$\dfrac{\alpha (\alpha + 1) (2 \alpha + 1)}{6} - \dfrac{\alpha ( \alpha + 1)}{2} + 1 - 1 = \dfrac{(\alpha + 1) * \alpha * (\alpha - 1)}{3} = \dfrac{( \alpha + 1)!}{3 * ( \alpha - 2)!}.$

Step 2: Show that x(1, n, 1) = (k + 1) / (r + 2).

$\displaystyle x(1,\ n,\ 1) = \sum_{m=2}^n \left \{ \dfrac{m - 1}{n - 1} * \dbinom{m}{1} * \dbinom{n - m}{1 - 1} \div \dbinom{n + 1}{1 + 1} \right \}\implies$

$\displaystyle x(1,\ n,\ 1) = \dfrac{1}{n - 1} * \sum_{m=2}^n \left \{ (m - 1) * m * \dbinom{n - m}{0} \div \dbinom{n + 1}{2} \right \} \implies$

$\displaystyle x(1,\ n,\ 1) = \dfrac{1}{n - 1} * \sum_{m=2}^n \left \{ (m^2 - m) * 1 * \dfrac{2! * \{(n + 1) - 2\}!}{(n + 1)!} \right \} \implies$

$\displaystyle x(1,\ n,\ 1) = \dfrac{1}{n - 1} * \sum_{m=2}^n \left \{ (m^2 - m) * \dfrac{2! * (n - 1)!}{(n + 1) * n * (n - 1)!} \right \} \implies$

$\displaystyle x(1,\ n,\ 1) = \dfrac{1}{n - 1} * \dfrac{2}{(n + 1) * n} * \sum_{m=2}^n (m^2 - m) \implies$

$\displaystyle x(1,\ n,\ 1) = \dfrac{2 * (n - 2)!}{(n + 1)!} * \sum_{m=2}^n (m^2 - m) \implies$

$x(1,\ n,\ 1) = \dfrac{2 * \cancel {(n - 2)!}}{\cancel {(n + 1)!}} * \dfrac{\cancel {(n + 1)!}}{3 * \cancel {(n - 2)!}} = \dfrac{2}{3} \implies$

$x(1,\ n,\ 1) = \dfrac{k + 1}{r + 1}.$

Now if we need induction, the next step would be to show that

$x(1,\ n,\ r) = \dfrac{k + 1}{r + 1}\ for\ 1 \le r \le (n - 1).$

We certainly have our base case for r = 1. But induction may not be the way to go.