I'm looking for a closed form expression for the sum

$P_n(x) =\sum_{0\leq k\leq n/2}\binom{n}{k}x^k$,

where $n$ is a given positive integer and $k$ runs over nonnegative integers between $0$ and $n/2$. The first such polynomials are


Alternatively one might want to subtract the half of the last term $\binom{n}{n/2}x^{n/2}$ when $n$ is even to obtain the polynomials

$Q_1(x) = 1, Q_2(x) = 1+x, Q_3(x) = 1+3x, Q_4(x) = 1 + 4x + 3x^2,\ldots$

These have the nice property that $Q_n(x) + x^n Q_n(1/x) = (1+x)^n$, but I am unable to obtain a closed form for either sequence of functions. One can obtain a recursive formula and also a closed form expression for the generating function

$G(x,y) = \sum_{n=0}^{\infty} Q_n(x)y^n$,

but this form involves a term of the type

$\frac{\sqrt{1- 4y}}{1 - ay},$

which I am unable to expand as a power series in $y$. If you find an expression for the sequences $P_n$ or $Q_n$ or are able to express the above function as a power series in $y$, I would be very grateful. Thanks!


$\sum_{0\leq k\leq n/2}\binom{n}{k}x^k = (x+1)^n - \binom{n}{\left\lfloor\frac{n}{2}\right\rfloor + 1} x^{\left\lfloor\frac{n}{2}\right\rfloor + 1} {_2F_1}\left(1,\left\lfloor\frac{n}{2}\right\rfloor - n + 1;\left\lfloor \frac{n}{2}\right\rfloor + 2;-x\right)$,

where $_2F_1$ is the hypergeometric function.


The power series for

$$f(y) = \sqrt{1-4y} \cdot \frac{1}{1 - ay} $$

is straightforward to compute: take the series for each factor separately and multiply.

$$\begin{align} f(y) = \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \binom{1/2}{i} (-4y)^i (ay)^j = \sum_{n=0}^{\infty} \left( \sum_{i=0}^n \binom{1/2}{i} (-4)^i a^{n-i} \right) y^n \end{align}$$

WolframAlpha can find a closed form expression for the coefficient on $y^n$, but it involves hypergeometric functions.

  • $\begingroup$ If there's a closed form expression for the coefficient of $y^n$ in terms of hypergeometric functions, then I guess the original polynomials can be expressed in terms of hypergeometric functions as well, which is pretty nice already. Thanks! $\endgroup$ – user62159 Feb 14 '13 at 9:50

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