According to Mathematica, your sum equals:

$(1-p^2)^{l/2} \mbox{LegendreP}\left(l/2, \frac{1+p^2}{1-p^2}\right),$

or

$\, _2F_1\left(-\frac{l}{2},-\frac{l}{2};1;p^2\right)$

The second sum is $n^2 \binom{2 n-2}{n-1}.$

Ain't technology grand...

EDIT The real question is: why do you want to know? The expressions I give above allow you to get asymptotics, get ODE satisfied by the functions, etc, etc. If you want an expression in terms of "elementary functions" (whatever that means in this case), with very high probability there are not any (this is less certain here because these are definite summmations. I strongly advise you to read A=B.

According to Mathematica, your sum equals:

$(1-p^2)^{l/2} \mbox{LegendreP}\left(l/2, \frac{1+p^2}{1-p^2}\right),$

or

$\, _2F_1\left(-\frac{l}{2},-\frac{l}{2};1;p^2\right)$

The second sum is $n^2 \binom{2 n-2}{n-1}.$

Ain't technology grand...

EDIT The real question is: why do you want to know? The expressions I give above allow you to get asymptotics, get ODE satisfied by the functions, etc, etc. If you want an expression in terms of "elementary functions" (whatever that means in this case), with very high probability there are not any (this is less certain here because these are definite summmations). I strongly advise you to read Petkovsek and Zeilberger's "A=B."

According to Mathematica, your sum equals:

$(1-p^2)^{l/2} \mbox{LegendreP}\left(l/2, \frac{1+p^2}{1-p^2}\right),$

or

$\, _2F_1\left(-\frac{l}{2},-\frac{l}{2};1;p^2\right)$

The second sum is $n^2 \binom{2 n-2}{n-1}.$

Ain't technology grand...

According to Mathematica, your sum equals:

$(1-p^2)^{l/2} \mbox{LegendreP}\left(l/2, \frac{1+p^2}{1-p^2}\right),$

or

$\, _2F_1\left(-\frac{l}{2},-\frac{l}{2};1;p^2\right)$

The second sum is $n^2 \binom{2 n-2}{n-1}.$

Ain't technology grand...

EDIT The real question is: why do you want to know? The expressions I give above allow you to get asymptotics, get ODE satisfied by the functions, etc, etc. If you want an expression in terms of "elementary functions" (whatever that means in this case), with very high probability there are not any (this is less certain here because these are definite summmations. I strongly advise you to read A=B.