Binomial Theorem/Abel's Generalisation/Negative n
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Theorem
Abel's Generalisation of Binomial Theorem:
- $\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$
does not hold for $n \in \Z_{< 0}$.
Proof
Putting $n = x = -1$ and $y = z = 1$ into the left hand side
- $\paren {-1 + 1}^{-1} = \dfrac 1 0$
which is undefined.
Putting the same values into the right hand side gives:
\(\ds \) | \(\) | \(\ds \sum_k \dbinom {-1} k \paren {-1} \paren {-1 - k}^{k - 1} \paren {1 + k}^{-1 - k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \paren {-1}^k \dbinom {-1} k \paren {1 + k}^{-2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 0} \paren {1 + k}^{-2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} \dfrac 1 {k^2}\) | Translation of Index Variable of Summation: Corollary | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\pi^2} 6\) | Basel Problem |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $52$