Centering \vdots in a system of many equations

\documentclass{article}
\usepackage{amsmath,calc}
\begin{document}

\begin{align*}
  R(-1) &= \sum_{i=1}^m A(i)R(i-1) \\
  R(-2) &= \sum_{i=1}^m A(i)R(i-2) \\
        &\mathrel{\makebox[\widthof{=}]{\vdots}} \\
  R(-k) &= \sum_{i=1}^m A(i)R(i-k)
\end{align*}

\end{document}

Consider the package mathtools, which provides several corrections for and additions to amsmath.

\usepackage{mathtools}

It also provides a comfortable solution for your problem. You can even choose between a normal (\vdotswithin) and a short (\shortvdotswithin) distance.

\begin{align*}
  a &= b \\
  & \vdotswithin{=} \\
  & = c \\
  \shortvdotswithin{=}
  & = d
\end{align*}

The result convinces.

More details can be found in the documentation of the package, section "Centered \vdots", where also the example above is taken from.

Here is another option (not very popular among TeXans):

\begin{eqnarray}
  R(-1) &=& \sum_{i=1}^m A(i)R(i-1) \\
  R(-2) &=& \sum_{i=1}^m A(i)R(i-2) \\
        &\vdots& \\
  R(-k) &=& \sum_{i=1}^m A(i)R(i-k)
\end{eqnarray}