Changing the default colours and styles of equations using tcolorbox

I've been experimenting with the tcolorbox to create some LaTeX documents.

It is a wonderful package but I am struggling to see how to change the default colours around the displayed equation given in the Theorem example e.g., how do I change, or removed, the red bounding box around the equation. For example, how could I change it to green?

\newtcbtheorem[auto counter,number within=section]{theo}%
  {Theorem}{fonttitle=\bfseries\upshape, fontupper=\slshape,
     arc=0mm, colback=blue!5!white,colframe=blue!60!white}{theorem}

\begin{theo}{Summation of Numbers}{summation}
  For all natural number $n$ it holds:
  \begin{equation}
  \tcbhighmath{\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}.}
  \end{equation}
\end{theo}

A screen shot of what is currently generated is shown below:

It would also be useful to be able to specify different colours etc., in different Theorems. That is, rather than re-defining the colours for all Theorems, is there a way to have, for example, the first theorem having a green bounding box, the next theorem having a black bounding box, and the next one no bounding box?

You can adjust the highlight math style:

\documentclass{article}

\usepackage[most]{tcolorbox}

\newtcbtheorem[auto counter,number within=section]{theo}%
  {Theorem}{fonttitle=\bfseries\upshape, fontupper=\slshape,
     arc=0mm, colback=blue!5!white,colframe=blue!60!white,
       every box/.style={
         highlight math style={
           colback=green!10!white,
           colframe=green
         }
       }
     }{theorem}
     
\newtcbtheorem[auto counter,number within=section]{proof}%
  {Proof}{fonttitle=\bfseries\upshape, fontupper=\slshape,
     arc=0mm, colback=blue!5!white,colframe=blue!60!white,
       every box/.style={
         highlight math style={
           colback=blue!10!white,
           colframe=blue
         }
       }
     }{proof}     

\begin{document}

\begin{theo}{Summation of Numbers}{summation}
  For all natural number $n$ it holds:
  \begin{equation}
  \tcbhighmath{\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}.}
  \end{equation}
\end{theo}

\begin{proof}{Summation of Numbers}{summation}
  For all natural number $n$ it holds:
  \begin{equation}
  \tcbhighmath{\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}.}
  \end{equation}
\end{proof}

\end{document}