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==Theorem==
The following formula holds:
$$(n+2\lambda)C_n^{\lambda}(x) = 2\lambda \left(C_n^{\lambda+1}(x)-xC_{n-1}^{\lambda+1}(x) \right),$$
where $C_n^{\lambda}$ denotes [[Gegenbauer C]].

==Proof==

==References==

[[Category:Theorem]]
[[Category:Unproven]]

Return to (n+2lambda)C n^(lambda)(x)=2lambda(C n^(lambda+1)(x)-xC (n-1)^(lambda+1)(x)).

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