Sum of r+k Choose k up to n
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Theorem
Let $r \in \R$ be a real number.
Then:
- $\ds \forall n \in \Z: n \ge 0: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
where $\dbinom r k$ is a binomial coefficient.
Proof
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition
- $\ds \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
Basis for the Induction
$\map P 0$ is true, as $\dbinom r 0 = 1 = \dbinom {r + 1} 0$ by definition.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P m$ is true, where $m \ge 0$, then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
- $\ds \sum_{k \mathop = 0}^m \binom {r + k} k = \binom {r + m + 1} m$
Then we need to show:
- $\ds \sum_{k \mathop = 0}^{m + 1} \binom {r + k} k = \binom {r + m + 2} {m + 1}$
Induction Step
This is our induction step:
\(\ds \sum_{k \mathop = 0}^{m + 1} \binom {r + k} k\) | \(=\) | \(\ds \sum_{k \mathop = 0}^m \binom {r + k} k + \binom {r + m + 1} {m + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom {r + m + 1} m + \binom {r + m + 1} {m + 1}\) | from the induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {r + m + 2} {m + 1}\) | Pascal's Rule |
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
$\blacksquare$
Also see
- Rising Sum of Binomial Coefficients, where the result is proved for integer $r$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $\text{E} \ (10)$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $13$