Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$

Question

Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$

---

This looks like an unusual hockey stick sum. Here are my attempts:

Method 1:

The sum is equivalent to

$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$

and I could evaluate neither of these.

Method 2:

$$S=\text{coefficient of $x^n$ in }:$$

$$2^n(1+x)^n\Bigg( 1+\frac{1+x}{2}+\left(\frac{1+x}{2}\right)^2+\cdots+\left(\frac{1+x}{2}\right)^n \Bigg)$$

$$=(1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right)$$

It looks like a heavy task to collect all the $x^n$ coefficients from this expression.

---

Im out of ideas. Any hint is appreciated.

From Putnam 2020, Q A2

Answer 1

Let $$ S_n:=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}} $$ Obviously $S_0=1$. Assume that equality $$S_{n-1}=1\tag1$$ is valid for some $n$. Then it is valid for $n+1$ as well: $$ \begin{align} S_n &=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}\\ &=\sum_{k=0}^n\frac{\binom{n+k-1}{k-1}+\binom{n-1+k}{k}}{2^{n+k}}\\ &=\frac12\sum_{k=0}^{n-1}\frac{\binom{n+k}{k}}{2^{n+k}}+\frac12\sum_{k=0}^n\frac{\binom{n-1+k}{k}}{2^{n-1+k}}\\ &=\frac12S_n-\frac{\binom{2n}{n}}{2^{2n+1}}+\frac12S_{n-1}+\frac{\binom{2n-1}{n}}{2^{2n}}\\ &=\frac12S_n+\frac12S_{n-1}\implies S_n=S_{n-1}\stackrel{I.H.}=1. \end{align} $$

Thus, by induction the equality $(1)$ is valid for all integer $n\ge0$.

Accordingly your sum is $2^{2n}S_n=4^n$.

Answer 2

In another way $$ \eqalign{ & S = \sum\limits_{r = 0}^n {2^{\,n - r} \left( \matrix{ n + r \cr r \cr} \right)} = \cr & = \sum\limits_{k = 0}^n {\left( \matrix{ 2n - k \cr n - k \cr} \right)2^{\,k} = } \quad (1)\cr & = \sum\limits_{0\, \le \,k} {\left( \matrix{ 2n - k \cr n - k \cr} \right)2^{\,k} } = \quad (2) \cr & = \sum\limits_{0\, \le \,k} {\left( \matrix{ 2n - k \cr n - k \cr} \right)\left( {1 + 1} \right)^{\,k} } = \sum\limits_{0\, \le \,k} {\sum\limits_{0\, \le \,j} {\left( \matrix{ 2n - k \cr n - k \cr} \right) \left( \matrix{ k \cr k - j \cr} \right)} } = \quad (3) \cr & = \sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,n - j} \sum\limits_{\left( {0\, \le } \right)\,k} {\left( \matrix{ - n - 1 \cr n - k \cr} \right)\left( \matrix{ - j - 1 \cr k - j \cr} \right)} } = \quad (4) \cr & = \sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,n - j} \left( \matrix{ - n - j - 2 \cr n - j \cr} \right)} = \quad (5) \cr & = \sum\limits_{0\, \le \,j} {\left( \matrix{ 2n + 1 \cr n - j \cr} \right)} = \sum\limits_{k = 0}^n {\left( \matrix{ 2n + 1 \cr k \cr} \right)} = \quad (6)\cr & = {1 \over 2}\sum\limits_{k = 0}^{2n + 1} {\left( \matrix{ 2n + 1 \cr k \cr} \right)} = 2^{\,2n} \quad (7) \cr} $$ where the steps are:

• 1. change index;
• 2. remove upper bound (it is implicit in the binomial);
• 3. split the $2$;
• 4. upper negation ( $\binom{n}{m}=(-1)^m \binom{m-n-1}{m}$ ) on both binomials;
• 5. convolution in $k$,
• 6. upper negation;
• 7. symmetry of the binomial .

Answer 3

$$(1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right)$$

You were on the right path, but you just didn't take the final step. Finish:

$$ \frac{(1+x)^n 2^{n+1} }{(1-x)} - \frac{(1+x)^{2n+1} }{(1-x)}$$

Now, I will introduce a 'nice' result( By the cauchy product rule):

$$ \frac{ P(x)}{1-x}= (\sum_i x^i) \sum_{i} a_i x^i = \sum_u c_ux^u$$

Where:

$$ c_u = \sum_{i=0}^u a_i \tag{1}$$

Hence, the applying the coefficient operator and use(1) while I'm at it:

$$ [x^n] \left[\frac{(1+x)^n 2^{n+1} }{(1-x)} - \frac{(1+x)^{2n+1} }{(1-x)} \right] = 2^{n+1} [x^n] \left[ \frac{(1+x)^n}{1-x} \right]- [x^n] \left[ \frac{(1+x)^{2n+1} }{1-x} \right]\\= 2^{n+1} \sum_{i=0}^n \binom{n}{i}-\sum_{i=0}^n \binom{2n+1}{i}$$

Now, by standard results we can work out the first binomial sum:

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

And, the tricky sum/one step more complicated sum is:

$$ S= \sum_{i=0}^n \binom{2n+1}{i} = \sum_{i=0}^n \binom{2n+1}{(2n+1)-i} = \sum_{i=0}^n \binom{2n+1}{(2n+1) - (n-i) } = \sum_{i=0}^n \binom{2n+1}{n+1+i}$$

This leads to:

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

Or,

$$ S= 2^{2n}$$

Put everything together and you have the answer of $4^n$.

For more information about cauchy product rule, you can check out my article about it here

Answer 4

Consider $$\left( \begin{matrix} n \\ k \\ \end{matrix} \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|=R}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}$$ so $$\sum\limits_{r=0}^{n}{{{2}^{n-r}}\left( \begin{matrix} n+r \\ r \\ \end{matrix} \right)}=\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\sum\limits_{r=0}^{n}{{{2}^{n-r}}}\frac{{{\left( 1+z \right)}^{n+r}}}{{{z}^{r+1}}}dz}=\\\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\frac{{{2}^{n+1}}{{\left( 1+z \right)}^{n}}}{z-1}-\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( z-1 \right){{z}^{n+1}}}dz}\\=\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( 1-z \right){{z}^{n+1}}}dz}$$

Where for convenience we've chosen a circular contour enclosing the origin with radius less than $1$. We can do this because directly after the summation the only contribution comes from a residue at $z=0$ (note there is no residue at $z=1$ in the second line). The final line above reflects this. Now to get the residue at $z=0$... $$res\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( 1-z \right){{z}^{n+1}}}=res\sum\limits_{m=0}^{\infty }{{}}\sum\limits_{k=0}^{\infty }{\left( \begin{matrix} 1+2n \\ k \\ \end{matrix} \right)\frac{1}{{{z}^{n+1-m-k}}}}\\=\sum\limits_{m=0}^{\infty }{\left( \begin{matrix} 1+2n \\ n-m \\ \end{matrix} \right)=}\sum\limits_{m=0}^{n}{\left( \begin{matrix} 1+2n \\ n-m \\ \end{matrix} \right)}=\sum\limits_{m=0}^{n}{\left( \begin{matrix} 1+2n \\ m \\ \end{matrix} \right)}={{4}^{n}}$$ hence $$\sum\limits_{r=0}^{n}{{{2}^{n-r}}\left( \begin{matrix} n+r \\ r \\ \end{matrix} \right)}={{4}^{n}}$$

Answer 5

In seeking to evaluate

$$S_n = \sum_{r=0}^n 2^{n-r} {n+r\choose r}$$

we find that it is

$$[z^n] \frac{1}{1-2z} \frac{1}{(1-z)^{n+1}} = \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} \frac{1}{1-2z} \frac{1}{(1-z)^{n+1}}.$$

We will use the fact that residues sum to zero, which requires the residue at $z=1/2$ and the residue at $z=1$ as well as the residue at infinity. The latter is zero by inspection, however . We get for the residue at $z=1/2$

$$-\frac{1}{2} \mathrm{Res}_{z=1/2} \frac{1}{z^{n+1}} \frac{1}{z-1/2} \frac{1}{(1-z)^{n+1}}$$

We obtain

$$- \frac{1}{2} 2^{n+1} 2^{n+1} = - 2 \times 4^n.$$

We also have for the residue at $z=1$

$$\mathrm{Res}_{z=1} \frac{1}{z^{n+1}} \frac{1}{1-2z} \frac{1}{(1-z)^{n+1}} \\ = \mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{n+1}} \frac{1}{-1-2(z-1)} \frac{1}{(1-z)^{n+1}} \\ = (-1)^n \mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{n+1}} \frac{1}{1+2(z-1)} \frac{1}{(z-1)^{n+1}}.$$

This is

$$(-1)^n \sum_{r=0}^n (-1)^r {n+r\choose r} (-1)^{n-r} 2^{n-r} = \sum_{r=0}^n 2^{n-r} {n+r\choose r} = S_n.$$

We have shown that $S_n - 2 \times 4^n + S_n = 0$ or

$$\bbox[5px,border:2px solid #00A000]{ S_n = 4^n.}$$

For the residue at infinity we get

$$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{n+1} \frac{1}{1-2/z} \frac{1}{(1-1/z)^{n+1}} = - \mathrm{Res}_{z=0} z^n \frac{1}{z-2} \frac{z^{n+1}}{(z-1)^{n+1}} \\ = - \mathrm{Res}_{z=0} z^{2n+1} \frac{1}{z-2} \frac{1}{(z-1)^{n+1}} = 0.$$

Answer 6

Consider the number of binary strings that are $(2n+1)$ digits long with at least $(n+1)$ digits equal to $1$. Let this number be $S$.

Method 1: Flip the $1$'s into $0$'s and vice versa. This covers all binary strings of length $2n+1$: so $2S=2^{2n+1}\implies S=2^{2n}$

Method 2: Consider the $(n+1)th$ $1$ digit. Suppose this is the $(n+1+r)$th digit of the string. Then this yields the desired sum by hockey stick.

So the answer is $\boxed{2^{2n}}$.

Answer 7

OP's second method can be made to work and it gives the simplest as in the following solution.

$$S=\text{Coefficient of $x^n$ in}~~ (1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right).$$ $$S=-\text{Coefficient of $x^n$}~ \text{in} \left((1+x)^{2n+1}(1-x)^{-1}-2^{n+1}(1-x)^{-1}\right)$$ $$S=-\text{Coefficient of $x^n$}~ \text{in} \left((1+x)^{2n+1}\sum_{k=0}^{\infty} x^k-2^{n+1}\sum_{k=0}^{\infty}x^k\right)$$ $$S=- \left(\sum_{k=0}^{n} {2n+1\choose k}-2^{n+1}\sum_{k=0}^{\infty}{n \choose k}\right)=-2^{2n}+2^{2n+1}=2^{2n}$$

Answer 8

A combinatorial proof.

Let $U=\{A\subseteq \{1,2,\dots,2n+1\}, |A|>n\}.$

We have that $|U|=2^{2n+1}/2=2^{2n}.$

Given an $A\in U,$ define $m(A)$ as the largest number such that $|A\cap\{1,2,\dots,n+r\}|=n.$ $m(A)$ can be anywhere from $0$ to $n,$ and we always have $m(A)+1\in A.$

Let $U_r=\{A\in U\mid m(A)=r\}.$ What is the size of $U_r?$ We can choose any $n$ elements from $1,2,\dots,n+r$, add the element $n+r+1,$ and then any subset of elements from the remaining $n-r$ elements. So $$|U_r|=\binom{n+r}{n}2^{n-r}.$$

and: $$4^{n}=2^{2n}=|U|=\sum_{r=0}^{n}|U_r|=\sum_{r=0}^{n}2^{n-r}\binom{n+r}n$$