Simplifying $\sum_{i=0}^n i^k\binom{n}{2i+1}$

Question

What is the formula for

\begin{eqnarray}\sum_{i=0}^n i^k\binom{n}{2i+1}?\end{eqnarray}

I tried to use the identity $$ \sum_{i=0}^ni(i-1)\cdots(i-p)\binom{n}{2i+1}=(n-p-2)(n-p-3)\cdots(n-2p-2)2^{n-2p-3} $$ but got into a mess. Any idea?

Answer 1

Using Stirling numbers of the second kind we have:

$$ i^k = \sum_{j=0}^{k}{k \brace j}(i)_j = \sum_{j=0}^{k}{k \brace j}i\cdot(i-1)\cdot\ldots\cdot(i-j+1)\tag{1}$$ then, since you know that: $$ \sum_{i=0}^{n}(i)_{p+1}\binom{n}{2i+1}=(n-p-2)_{p+1} 2^{(n-1)-2(p+1)}\tag{2}$$ you have:

$$\begin{eqnarray*} \sum_{i=0}^{n}i^k\binom{n}{2i+1}&=&\sum_{i=0}^{n}\sum_{j=0}^{k}{k \brace j}(i)_j\binom{n}{2i+1}\\&=&\sum_{j=0}^{k}{k \brace j}(n-j-1)_j\, 2^{n-(2j+1)}.\tag{3}\end{eqnarray*} $$

On the other hand, your sum is a value of a Fibonacci polynomial, by definition.

Answer 2

The first step is to find the series evaluation of $$S_{n}(t) = \sum_{i=0}^{n} \binom{n}{2i+1} \, t^{i}.$$ One this is done utilize the operator $\delta = t \, \frac{d}{dt} = t \, D$ to obtain the desired series. This will be of the form $$S(1) = \left. \delta^{k} \, S_{n}(t) \right|_{t=1} = \sum_{i=0}^{n} i^{k} \, \binom{n}{2i+1}.$$

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$$S_{n}(t) = \frac{(1 + \sqrt{t})^{n} - (1 - \sqrt{t})^{n}}{2 \, \sqrt{t}}$$

It may be best to consider two functions in the process as well: \begin{align} \theta_{n}(t) &= \frac{(1 + \sqrt{t})^{n} - (1-\sqrt{t})^{n}}{\sqrt{t}} \\ \phi_{n}(t) &= (1+\sqrt{t})^{n} + (1 - \sqrt{t})^{n} \end{align}

Answer 3

Suppose we seek to evaluate $$S(n, k) = \sum_{q=0}^{n} q^k {n\choose 2q+1}.$$

Introduce $$q^k = \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(qz)}{z^{k+1}} \; dz.$$

Observe that with $k\ge 1$ we also get the correct value for $q=0.$

We obtain for the sum $$\frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+1}} \sum_{q=0}^{n} {n\choose 2q+1} \exp(qz) \; dz \\ = \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+1}} \sum_{q=0}^{n} {n\choose 2q+1} \exp(2q(z/2)) \; dz \\ = \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+1}} \exp(-z/2) \sum_{q=0}^{n} {n\choose 2q+1} \exp((2q+1)(z/2)) \; dz.$$

This is $$\frac{1}{2} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+1}} \exp(-z/2) ((1+\exp(z/2))^n-(1-\exp(z/2))^n) \; dz.$$

Substituting $z=2w$ we obtain $$\frac{1}{2^{k+1}} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w) ((1+\exp(w))^n-(1-\exp(w))^n) \; dw.$$

This has two pieces call them $A_1$ and $A_2.$ The piece $A_1$ is $$\frac{1}{2^{k+1}} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w) (1+\exp(w))^n\; dw \\ = \frac{1}{2^{k+1}} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w) \sum_{q=0}^n {n\choose q} 2^{n-q} (\exp(z)-1)^q \; dw.$$

This is by convolution of generating functions $$\frac{1}{2^{k+1}} \sum_{q=0}^n {n\choose q} 2^{n-q} q! \sum_{p=0}^k {k\choose p} (-1)^{k-p} {p\brace q}.$$

The piece $A_2$ is $$-\frac{1}{2^{k+1}} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w) (1-\exp(w))^n \; dw \\ = (-1)^{n+1} \frac{1}{2^{k+1}} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{w^{k+1}} \exp(-w) (\exp(w)-1)^n \; dw.$$

This is $$(-1)^{n+1} \frac{1}{2^{k+1}} n! \sum_{p=0}^k {k\choose p} (-1)^{k-p} {p\brace n}.$$

Observe that this vanishes if $k\lt n.$

Finally note concerning $A_1$ that if $k\lt n$ all the terms with $q\gt k$ from the sum in $q$ cease to contribute owing to the Stirling number being zero, so we may set the upper limit to $k$ in this case. Similarly when $n\lt k$ the terms with $q\gt n$ cease to contribute owing to the first binomial coefficient.

The end result is a polynomial in $n$ plus a correction term and we get $$\frac{1}{2^{k+1}} \sum_{q=0}^k {n\choose q} 2^{n-q} q! \sum_{p=0}^k {k\choose p} (-1)^{k-p} {p\brace q} + (-1)^{n+1} \frac{1}{2^{k+1}} n! \sum_{p=0}^k {k\choose p} (-1)^{k-p} {p\brace n}.$$