# How to evaluate the finite sum $\sum_{k=0}^n{\alpha \choose k}^2\lambda^k$

Is there a close form expression for the series $$\sum_{k=0}^n{\alpha \choose k}^2\lambda^k,\quad \alpha ~ \text{is non-integer}$$

As far as I know, there is an identity involving binomial coefficients $$\sum_{k=0}^n{n \choose k}^2 = {2n \choose n}$$ which can be proved using generating function $(1+x)^n(1+x)^n=(1+x)^{2n}$. However similar method can not be applied to the first series.

Is there a way to deal with it ? Thank you !

• I suppose this has to do with gamma function. – Troy Woo Aug 31 '14 at 9:03
• According to WA the closed form is $$(\lambda+1)^\alpha-\dfrac{\Gamma(a+1)\lambda^{n+1}\,_2F_1(1,-\alpha+n+1;n+2;- \lambda)}{\Gamma(n+2)\Gamma(\alpha-n)}$$ where $_2F_1(a,b;c;z)$ is the hypergeometric function – Alice Ryhl Aug 31 '14 at 10:03
• @Darksonn Can you say more about it? – ecook Aug 31 '14 at 10:05
• @ecook Not really, $\Gamma(x)$ is the gamma function which is related to factorials, namely for integer $x$ it is true that $x!=\Gamma(x+1)$ – Alice Ryhl Aug 31 '14 at 10:08
• Solutions based on hypergeometric functions ${}_2F_1$ are really rewritings of the original formula. – Did Aug 31 '14 at 10:38

A standard expression for the $$n$$th Legendre polynomial is $$\frac{1}{2^n}\sum_0^n{n\choose k}^2(x-1)^{n-k}(x+1)^k$$.
So that gives immediately the expression $$(1-x)^nP_n\left(\frac{1+x}{1-x}\right)$$
• This is only true for the $\alpha=n$ case of the OP---which, since the OP asks explicitly for $\alpha$ non-integer, is not the question at hand. – Semiclassical Aug 31 '14 at 17:31