$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}