Product of Number by its Falling Factorial
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Theorem
Let $x^{\underline n}$ denote the $n$th falling factorial power of $x$.
Then:
- $x x^{\underline n} = x^{\underline {n + 1} } + n x^{\underline n}$
Proof
\(\ds x x^{\underline n}\) | \(=\) | \(\ds \paren {x - n + n} x^{\underline n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x - n} x^{\underline n} + n x^{\underline n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x - n} \prod_{j \mathop = 0}^{n - 1} \paren {x - j} + n x^{\underline n}\) | Definition of Falling Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{j \mathop = 0}^n \paren {x - j} + n x^{\underline n}\) | Definition of Continued Product | |||||||||||
\(\ds \) | \(=\) | \(\ds x^{\underline {n + 1} } + n x^{\underline n}\) | Definition of Falling Factorial |
$\blacksquare$