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:

\(\ds \sum_k \paren {-1}^k \dbinom 0 k \dbinom {r - k t} 0 \dfrac r {r - k t}\)

\(=\)

\(\ds \sum_k \paren {-1}^k \delta_{0 k} \dbinom {r - k t} 0 \dfrac r {r - k t}\)

Zero Choose n

\(\ds \)

\(=\)

\(\ds \dbinom r 0 \dfrac r r\)

All terms but for $k = 0$ vanish

\(\ds \)

\(=\)

\(\ds 1\)

Binomial Coefficient with Zero

\(\ds \)

\(=\)

\(\ds \delta_{0 0}\)

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:

\(\ds \)

\(\)

\(\ds \sum_k \paren {-1}^k \dbinom {j + 1} k \dbinom {r - k t} {j + 1} \dfrac r {r - k t}\)

\(\ds \)

\(=\)

\(\ds \sum_k \paren {-1}^k \paren {\dbinom j k + \dbinom j {k - 1} } \dbinom {r - k t} {j + 1} \dfrac r {r - k t}\)

Definition of Binomial Coefficient

This needs considerable tedious hard slog to complete it. In particular: brain not working tonight. Ought to be able to split binomial coeffs and reduce but it didn't do what it should. Perhaps try to solve it by reducing to factorials. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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

Therefore:

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