$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Thanks to a comment of $\ds{\tt@JimmyK_{4542}}$, I found a missing term in a previous calculation. Indeed, the result turns out to be very simple:
\begin{align}
&\color{#66f}{\large\sum_{k = 1}^{41}{83 \choose k}}
=\half\bracks{%
\sum_{k = 1}^{41}{83 \choose k} + \sum_{k = 1}^{41}{83 \choose 83 - k}}
=\half\bracks{%
\sum_{k = 1}^{41}{83 \choose k} + \sum_{k = -82}^{-42}{83 \choose -k}}
\\[3mm]&=\half\bracks{%
\sum_{k = 1}^{41}{83 \choose k} + \sum_{k = 82}^{42}{83 \choose k}}
=\half\bracks{%
\sum_{k = 0}^{83}{83 \choose k} - {83 \choose 0} - {83 \choose 83}}
=\half\pars{2^{83} - 2}
\\[3mm]&=\color{#66f}{\Large 2^{82} - 1}
\end{align}