Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt

Theorem

 * $\displaystyle \sum_k \left({-1}\right)^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$

where $\delta_{n 0}$ is the Kronecker delta.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_k \left({-1}\right)^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$

Basis for the Induction
$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

Induction Hypothesis
Now it needs to be shown that, if $P \left({j}\right)$ is true, where $j \ge 0$, then it logically follows that $P \left({j + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_k \left({-1}\right)^k \dbinom j k \dbinom {r - k t} j \dfrac r {r - k t} = \delta_{j 0}$

from which it is to be shown that:
 * $\displaystyle \sum_k \left({-1}\right)^k \dbinom {j + 1} k \dbinom {r - k t} {j + 1} \dfrac r {r - k t} = \delta_{\left({j + 1}\right) 0}$

Induction Step
This is the induction step:

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0} \sum_k \left({-1}\right)^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$