Computing a sum of binomial coefficients: $\sum_{i=0}^m \binom{N-i}{m-i}$ Does anyone know a better expression than the current one for this sum?

$$
\sum_{i=0}^m \binom{N-i}{m-i}, \quad 0 \le m \le N.
$$

It would help me compute a lot of things and make equations a lot cleaner in the context where it appears (as some asymptotic expression of the coefficients of a polynomial defined over cyclic graphs). Perhaps the context doesn't help much though.
For instance, if $N = m$, the sum is $N+1$, and for $N = m+1$, this sum is $N + (N-1) + \dots + 1 = N(N+1)/2$. But otherwise I don't know how to compute it. 

First note that $\dbinom{n-i}{m-i} = \dbinom{n-i}{n-m}$. Hence, you have
$$\dbinom{n-m}{n-m}+\dbinom{n-m+1}{n-m}+\cdots+\dbinom{n-1}{n-m}+\dbinom{n}{n-m}$$
Let $n-m = k$. Hence, we want to find an expression for
$$\dbinom{k}k+\dbinom{k+1}{k}+\cdots+\dbinom{n-1}{k}+\dbinom{n}{k}$$
From the Pascal's triangle, we have
$\hskip2.5in$
\begin{align}
\color{red}{\dbinom{n+1}{k+1}} & = \color{blue}{\dbinom{n}{k}} + \color{red}{\dbinom{n}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{red}{\dbinom{n-1}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \color{red}{\dbinom{n-2}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{red}{\dbinom{k+2}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{red}{\dbinom{k+1}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{blue}{\dbinom{k}{k}}
\end{align}
Hence, your result is $\dbinom{n+1}{n-m+1} = \dbinom{n+1}m$
