Formula:KLS:09.13:05

{\displaystyle{\displaystyle{\displaystyle x{\widehat{y}}_{n}\!\left(x\right)=% {\widehat{y}}_{n+1}\!\left(x\right)-\frac{2a}{(2n+a)(2n+a+2)}{\widehat{y}}_{n}% \!\left(x\right){}-\frac{4n(n+a)}{(2n+a-1)(2n+a)^{2}(2n+a+1)}{\widehat{y}}_{n-% 1}\!\left(x\right)}}}

Proof

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