How do I calculate the following sum $\sum_{i=0}^{n+1} \dbinom{n+1-i}{i} $

Question

$$\sum_{i=0}^{n+1} \dbinom{n+1-i}{i} =\ ?$$ I tried to find out the coefficient of $x^{n+1}$ in the formula $(1+x)^{n+1}(1+x+\dots+x^{n+1})$ but failed to find out the next step.I'm also looking for a combinatorial explaination of the sum.

Answer 1

We might as well get the formula with $n+1$ replaced by $n$.

Denote by $F_n$ the $n$-th Fibonacci number, that is, $$F_0:=1,\quad F_1:=1$$ and for $n\geq 0$, put $$F_{n+2}:=F_{n+1}+F_n.$$ I claim that $$\sum_{k}\binom{n-k}{k}=F_n,$$ where $k$ runs through all integers (with the usual convention of $\binom{a}{b}:=0$ in case $b<0$ or $b>a$).

We can prove this by strong induction. The cases $n=0$ and $n=1$ are immediate. Next, for $n\geq 0$, we need to show that $$\sum_k\binom{n+2-k}{k}=F_{n+2}.$$

We have $$\sum_{k}\binom{n+2-k}{k}=\sum_k\binom{n+1-k}{k}+\sum_k\binom{n+1-k}{k-1}$$ using the binomial theorem.

A change of variable $l:=k-1$ in the second sum gives $$\sum_{k}\binom{n+2-k}{k}=\sum_k\binom{n+1-k}{k}+\sum_l\binom{n-l}{l}=F_{n+1}+F_n=F_{n+2},$$ the second equality applying the induction hypothesis, and the last equality using the definition of the $F_n$.

Edit: The sum we are evaluating can be represented as shown here.

Answer 2

Benjamin and Quinn give a combinatorial proof using the manifestation of the Fibonacci numbers as tilings of a length $n$ board by squares and dominos, i.e., integer compositions with parts 1 and 2; see Identity 4 of Proofs That Really Count (MAA, 2003).

Let me give another combinatorial argument using a result of Cayley that compositions with no parts 1 are also counted by the Fibonacci numbers.

To see the connection to the Fibonacci numbers, to make Cayley's compositions of 6, namely $\{6, 4+2, 3+3, 2+4, 2+2+2\}$, add a 2 at the end of the such compositions of 4, so $\{4, 2+2\}$ gives $\{4+2, 2+2+2\}$, and extend by 1 the last of part of compositions of 5, so $\{5, 3+2, 2+3\}$ gives $\{6, 3+3, 2+4\}$. With the indexing $F_0 = 1$, $F_1 = 1$, the number of Cayley compositions of $n$ is $F_{n-1}$.

Now for your sum. As MacMahon explained, binomial coefficients count compositions with a particular number of parts. Specifically, $\binom{n+1-i}{i}$ counts the compositions of $n+2-i$ with $i+1$ parts. To connect the compositions counted by $\sum_i \binom{n+1-i}{i}$ to Cayley's compositions, increase each part by 1. This makes exactly the compositions of $n+3$ with no parts 1 and there are $F_{n+2}$ of those.

As an example, the sum $\binom{4}{0} + \binom{3}{1} + \binom{2}{2}$ (your $n=3$ case) corresponds to the compositions $5$; $3+1, 2+2, 1+3$; $1+1+1$. Increasing each part by 1 gives $6, 4+2, 3+3, 2+4, 2+2+2$, the $F_5 = 5$ Cayley compositions of 6.