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

Theorem

 * $\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$

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

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$

Basis for the Induction
$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\ds \sum_k \paren {-1}^k \dbinom j k \dbinom {r - k t} j \dfrac r {r - k t} = \delta_{j 0}$

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

Induction Step
This is the induction step:

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

Therefore:
 * $\ds \forall n \in \Z_{\ge 0} \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$