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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!

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It's called sum of rows of Pascal triangle, which does not exist in closed form. Try reading an article by Tomas Worsch (1994) called 'Lower and upper bounds for (sums of) binomial coefficients'. See also these two questions:…,… – Alex Feb 13 '13 at 14:37

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

where $_2F_1$ is the hypergeometric function.

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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.

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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! – user62159 Feb 14 '13 at 9:50

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