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From: Bob Hanlon on 3 Jun 2010 05:39

Clear[p];

p[x_, n_] = (-1 + x)^(n + 1)*
Sum[Binomial[n + 2*k, 2*k]*x^k, {k, 0, Infinity}] // FullSimplify

((1/2)*(-(1 - Sqrt[x])^(n + 1) -
(Sqrt[x] + 1)^(n + 1))*
(x - 1)^n)/(1 - x)^n

Simplify[(1 - x)^-n (x - 1)^n == (-1)^n,
Element[n, Integers]]

True

p[x_, n_Integer] = ((1/2)*(-(1 - Sqrt[x])^(n + 1) -
(Sqrt[x] + 1)^(n + 1)))*
(-1)^n;


Bob Hanlon

---- Roger Bagula <roger.bagula(a)gmail.com> wrote:

=============
The zero and one terms for the binomial version exist,
but this Narayana number analog has difficulty in Mathematica on the
n=1 term :
Clear[p, x]
p[x_, n_] = (1 - x)^(2*n - 1)*Sum[(Binomial[2*k + n, 2*k] Binomial[2*k
+
n, 1 + 2*k]/(2*k + n))*x^k, {k, 0, Infinity}]
Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0,
10}]
It is probably, like the n equal zero term, one.
Try :
Clear[p, x]
p[x_, n_] = (-1 + x)^(
n + 1)*Sum[Binomial[n + 2*k, 2*k]*x^k, {k, 0, Infinity}]
Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0,
10}]
( see if you can find the really simple Sqrt[x] form for this!)

I wouldn't have any chance doing these without Mathematica.
It also fails several ways on the Eulerian numbers analogs
of these infinite sums which are interesting in what they might look
like.

Respectfully, Roger L. Bagula
11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
http://www.google.com/profiles/Roger.Bagula
alternative email: roger.bagula(a)gmail.com


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