Evaluating Combination Sum $\sum{n+k\choose 2k} 2^{n-k}$

Question

Evaluate $$\sum_{k=0}^n{n+k\choose 2k} 2^{n-k}$$ So im not really sure how to begin with this. I would imagine we start with dividing out $2^{n}$, but not really sure much past that

Answer 1

The method used here is that of the generating function. Let $S_{n}$ be the series to be summed \begin{align} S_{n} = \sum_{k=0}^{n} \binom{n+k}{2k} \ 2^{n-k}. \end{align} The generating function and its reduction are as follows. \begin{align} \sum_{n=0}^{\infty} S_{n} \frac{t^{n}}{2^{n}} &= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{n+k}{2k} \ 2^{n-k} \ \frac{t^{n}}{2^{n}} \\ &= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \binom{n+2k}{2k} \ 2^{-k} t^{n+k} \\ &= \sum_{k=0}^{\infty} \frac{t^{k}}{2^{k}} \cdot \sum_{n=0}^{\infty} \frac{(2k+1)_{n} t^{n}}{n!} \\ &= \sum_{k=0}^{\infty} \frac{t^{k}}{2^{k}} \cdot (1-t)^{-2k-1} \\ &= \frac{1}{1-t} \sum_{k=0}^{\infty} \left( \frac{t}{2(1-t)^{2}} \right)^{k} = \frac{1-t}{1 - (5/2)t + t^{2}}. \end{align}

Now, \begin{align} \frac{1-t}{1 - (5/2)t + t^{2}} &= \frac{1-t}{(1/2-t)(2-t)} = \frac{2}{3} \left[ \frac{1}{1-2t} + \frac{1}{2(1-t/2)} \right] \\ &= \sum_{n=0}^{\infty} \left[ \frac{2^{2n+1}+1}{3 \cdot 2^{n}} \right] \ t^{n} \end{align}
and \begin{align} \sum_{n=0}^{\infty} S_{n} \frac{t^{n}}{2^{n}} = \sum_{n=0}^{\infty} \left[ \frac{2^{2n+1}+1}{3 \cdot 2^{n}} \right] \ t^{n} \end{align} which yields \begin{align} \sum_{k=0}^{n} \binom{n+k}{2k} \ 2^{n-k} = \frac{2^{2n+1}+1}{3}. \end{align}

Answer 2

Suppose we seek to evaluate $$\sum_{k=0}^n {n+k\choose 2k} 2^{n-k}$$ inspired by the work at this MSE link I and this MSE link II.

Start from $${n+k\choose 2k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2k+1}} (1+z)^{n+k} dz.$$

This yields the following expression for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \sum_{k=0}^n \frac{1}{z^{2k+1}} (1+z)^{n+k} \times 2^{n-k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} 2^n \times \frac{(1+z)^n}{z} \sum_{k=0}^n \left(\frac{1+z}{2z^2}\right)^k \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} 2^n \times \frac{(1+z)^n}{z} \frac{\left(\frac{1+z}{2z^2}\right)^{n+1} - 1} {\frac{1+z}{2z^2} - 1} \; dz \\= \frac{1}{2\pi i} \int_{|z|=\epsilon} 2^n \times \frac{(1+z)^n}{z \times (2z^2)^{n+1}} \frac{(1+z)^{n+1} - (2z^2)^{n+1}} {\frac{1+z}{2z^2} - 1} \; dz \\= \frac{1}{2\pi i} \int_{|z|=\epsilon} 2^n \times \frac{(1+z)^n}{z \times (2z^2)^n} \frac{(1+z)^{n+1} - (2z^2)^{n+1}} {1+z - 2z^2} \; dz \\= \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{2n+1}} \frac{(1+z)^{n+1} - (2z^2)^{n+1}} {1+z - 2z^2} \; dz.$$

It now follows from the Cauchy Residue Theorem that the integral is given by $$[z^{2n}] \left((1+z)^n \times \frac{(1+z)^{n+1} - (2z^2)^{n+1}} {1+z - 2z^2}\right).$$

The term in $(2z^2)^{n+1}$ starts at $z^{2n+2}$ and hence does not contribute. This leaves $$[z^{2n}] \frac{(1+z)^{2n+1}}{1+z - 2z^2} = [z^{2n}] (1+z)^{2n+1} \left(\frac{2}{3}\frac{1}{1+2z} +\frac{1}{3}\frac{1}{1-z}\right).$$

This finally yields for the coefficient the value $$\sum_{q=0}^{2n} \left( [z^{2n-q}] \frac{1}{1+z - 2z^2} \right) {2n+1\choose q} \\ = \sum_{q=0}^{2n} {2n+1\choose q} \frac{2}{3} (-2)^{2n-q} + \sum_{q=0}^{2n} {2n+1\choose q} \frac{1}{3} \times 1^{2n-q}.$$

This is $$\frac{2^{2n+1}+1}{3}.$$

Quite a remarkable method and suitable for implementation by a CAS. (The algebra is very simple even if it may appear challenging at first sight.)

Addendum Aug 2023. This is straightforward using Egorychev method, see below. Rather than cancel the above we leave it for reference as it represents one the first instances of the method appearing at MSE. This is the streamlined version:

Starting from

$$\sum_{k=0}^n {n+k\choose 2k} 2^{n-k} = \sum_{k=0}^n {n+k\choose n-k} 2^{n-k} \\ = 2^n [z^n] (1+z)^n \sum_{k\ge 0} z^k (1+z)^k 2^{-k}.$$

Here we have extended to infinity due to the coefficient extractor. Continuing,

$$2^n [z^n] (1+z)^n \frac{1}{1-z(1+z)/2} \\ = 2^n \; \underset{z}{\mathrm{res}}\; \frac{1}{z^{n+1}} (1+z)^n \frac{1}{1-z(1+z)/2}.$$

Now put $z/(1+z)=w$ so that $z=w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$ to get

$$2^n \; \underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1-w) \frac{1}{1-w/(1-w) / (1-w) /2} \frac{1}{(1-w)^2} \\ = 2^n \; \underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \frac{1-w}{(1-w)^2-w/2} \\ = 2^n \; \underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \frac{1}{3} \left[\frac{1}{1-w/2} + \frac{2}{1-2w}\right] \\ = 2^n \frac{1}{3} \left[ \frac{1}{2^n} + 2^{n+1} \right] = \frac{2^{2n+1} + 1}{3}.$$

The generating function is the same as that of @Leucippus.

Answer 3

Let's rewrite our sum $S(2n)$ as $\sum_l\binom{2n-l}l2^l$.

Recall that $\binom{2n-l}l$ is the number of tilings of a rectangle $1\times 2n$ by $l$ dominoes and $2n-2l$ squares (we choose which of $2n-l$ tiles are dominoes). So $S(2n)$ counts the number of tiling of a rectangle $1\times 2n$ by squares and two kinds of dominoes.

So obviously $S(N)=S(N-1)+2S(N-2)$ (cf. Fibonacci numbers). This linear recurrence can be solved using standard methods.

Answer 4

Both Marko Riedel and Leucippus gave algebraic proofs that the sum equals $(2^{2n+1}+1)/3$. Then, Grigory M showed that your sum equals the number of ways to tile a $1\times 2n$ rectangle with one variety of square and two varieties of domino. The purpose of my answer is to give a combinatorial proof for why the number of tilings is $(2^{2n+1}+1)/3$. To make the pattern more clear, I will write this as $$ \frac{2\cdot 2^{2n}+1}{3} $$ That is, if you look at binary sequences of length $2n$, then approximately $2/3$ of them should correspond to tilings with one type of square and two types of domino.

Given a tiling of a $1\times (2n)$ board by squares and two types of dominoes, we can produce a binary sequence of length $2n$ by making the following replacements: $$ \text{square}\to 0\qquad \text{domino}_1\to 10\qquad \text{domino}_2\to 11 $$ Conversely, given binary sequence, how can we tell whether or not it can generated by some tiling using this procedure? With a little effort, you can prove the following:

• Any binary sequence ending with $01$ does not correspond to a tiling. This is because every "$1$" in a tiling sequence is paired with a $1$ to its left, or a $1$ or $0$ to its right, but none of these are possible. There are $2^{2n-2}$ such sequences.

• Any binary sequence ending with $0111$ does not correspond to a tiling (the middle $1$ must pair with the right $1$, but then the left $1$ has nothing to pair with). There are $2^{2n-4}$ such sequences.

• Any binary sequence ending with $011111$ does not correspond to a tiling. There are $2^{2n-6}$ such sequences.

Continuing in this fashion, the number of bad sequences is $$ 2^{2n-2}+2^{2n-4}+2^{2n-6}+\dots+1=\frac{2^{2n}-1}3=2^{2n}-\frac{2\cdot 2^{2n}+1}{3} $$ This completes the proof that the number of good sequences is $(2\cdot 2^{2n}+1)/3$

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We can make this a little more combinatorial. Given an arbitrary binary sequence, $s$, define a new binary sequence, $\sigma(s)$, as follows:

• If the last two symbols of $s$ are not equal to $11$, then $\sigma(s)$ is defined by replacing the last two symbols of $s$ using the following rule: $$ 00\mapsto 10,\qquad 10\mapsto 01,\qquad 01\mapsto 00 $$

• Suppose instead that the two symbols of $s$ are equal to $11$. Letting $t$ be the remainder when this $11$ is removed from the end, then $\sigma(s)=\sigma(t)+11$, where $+$ denotes concatenation of sequences.

• The previous bullet point is recursive, so $\sigma$ needs a base case. We define $\sigma(11)=11$. This implies that $\sigma(11\cdots 1)=11\dots 1$.

You can show that $\sigma$ generates a $(\mathbb Z/3\mathbb Z)$ action on the set of $2^{2n}$ binary sequences. This action has exactly one fixed point, the all-ones sequence, so all other orbits have size $3$. You can prove that, in each nontrivial orbit, exactly one sequence does not correspond to a tiling. Specifically, each orbit has three sequences that end with $2k$ ones, for some number $k$. Deleting these ones, the three sequences will end with $00$, $01$, and $10$. The sequence ending in $01$ is the one that does not correspond to a tiling.

We have partitioned the set of $2^{2n}-1$ sequences (everything except $1\dots 1$) into groups of $3$, where two out of each group of three corresponds to a tiling. The all ones sequences is a valid tiling as well. Therefore, the number of tilings is $$ \frac23\left({2^{2n}-1}\right)+1, $$ which equals the desired expression.