Best ways to calculate sums in LaTeX

it is easier to use lualatex and luacode. For example

\DocumentMetadata{}
\documentclass[margin=5pt, varwidth]{standalone}
\usepackage{amsmath}
\usepackage{tikz}
\usepackage{luacode}
\begin{luacode}
function LuaSum(a,b,f)
  k = 0
  for i=a,b do
    k = k + f(i)
  end
  return k
end

function Binomial( n, k )
  if k > n then return nil end
  if k > n/2 then k = n - k end       --   (n k) = (n n-k)
  numer, denom = 1, 1
  for i = 1, k do
    numer = numer * ( n - i + 1 )
    denom = denom * i
  end
  return numer / denom
end
\end{luacode}
\begin{document}
Test 1: $\displaystyle\sum\limits_{k=1}^{100} k = \directlua{%
  tex.print(LuaSum(1, 100,  function(k) return k end ))}$

Test 2: $\displaystyle\sum\limits_{k=0}^{5} \dbinom{5}{k} = 
  \directlua{tex.print(LuaSum(0, 5,  function(k) return Binomial(5,k) end ))}$
\end{document}

With return numer // denom you'll get the integer 32 instead of the floating number ...

Have you tried the package numerica ?

\documentclass[margin=5pt, varwidth]{standalone}
\usepackage{amsmath}
\usepackage{numerica}

\begin{document}

Test 1:
\[
\sum_{k=1}^{100} k
= \eval*{\sum_{k=1}^{100} k}
\]

Test 2:
\[
\sum_{k=0}^{5} \binom{5}{k}
= \eval*{\sum_{k=0}^{5} \binom{5}{k}}
= 2^5
\]

\end{document}

Edit 1 with a table of values

\documentclass[margin=5pt, varwidth]{standalone}

\usepackage{amsmath}
\usepackage{numerica}
\usepackage{numerica-tables}

\begin{document}

% Table of values for S(n) = \sum_{k=1}^{n} k, for n={1,..,10}
% Key point:
% - rvar is the row variable
% - rstop is how far we go
% - [n=1] gives the initial value of the row variable

\begin{table}[h]
\centering
\nmcTabulate[
  rvar=n,
  rstop=10,
  rstep=1
]{ \sum_{k=1}^{n} k }[n=1]

\end{table}

\end{document}

You can use xintexpr package

\documentclass{article}
\usepackage{xintexpr}
\newcommand{\myfrac}[1]{\xintTeXsignedFrac{\xinteval{reduce(#1)}}}
\usepackage{amsmath}
\begin{document}
\[1 + 2 + \cdots + 100=\myfrac{add(x, x = 1..100)}.\]
\[1 + \dfrac{1}{2} + \cdots + \dfrac{1}{50}=\myfrac{add(1/x, x = 1..50)}.\]
\[1 + \dfrac{1}{2^2} + \cdots + \dfrac{1}{50^2}=\myfrac{add(1/x^2, x = 1..50)}.\]
\end{document} 

I tried with Mathematica

^{I will delete this later. Probably.}

\documentclass[margin=5pt, varwidth]{standalone}
\usepackage{amsmath}
\ExplSyntaxOn
\cs_new:Npn \cis_eval_sum:n #1 {}
\cs_new_protected:Npn \cis_eval_real:nnnn #1#2#3#4 
{
  \fp_zero:N \l_tmpa_fp
  \cs_set:Npn \cis_eval_sum:n ##1 
  {
    \fp_set:Nn #2 {##1}
    \fp_add:Nn \l_tmpa_fp {#1}
  }
  \int_step_function:nnN {#3}{#4} \cis_eval_sum:n
  \fp_to_decimal:N \l_tmpa_fp
}
\cs_new_protected:Npn \cis_eval_ints:nnnn #1#2#3#4 
{
  \int_zero:N \l_tmpa_int
  \cs_set:Npn \cis_eval_sum:n ##1 
  {
    \cs_set:Npn #2 {##1}
    \int_add:Nn \l_tmpa_int {#1}
  }
  \int_step_function:nnN {#3}{#4} \cis_eval_sum:n
  \int_to_arabic:n {\l_tmpa_int}
}
\NewDocumentCommand \Sum { smmmm }
{
  \group_begin:
  \IfBooleanTF {#1}
  {
    \cis_eval_ints:nnnn {#2}{#3}{#4}{#5}
  }{                                   
    \cis_eval_real:nnnn {#2}{#3}{#4}{#5}
  }
  \group_end:
}
\ExplSyntaxOff
\begin{document}
Test 1: $\displaystyle\sum\limits_{k=1}^{100} k = \Sum{\k}{\k}{1}{100}$

\newcommand\binomial[2]{\fpeval{fact(#1)/(fact(#2)*fact(#1-#2))}}
Test 2: $\displaystyle\sum\limits_{k=0}^{5} \dbinom{5}{k}
= \Sum{  \binomial{5}{\k}  }{\k}{0}{5} =2^5 = \fpeval{2^5}$

Test 2: $\displaystyle\sum\limits_{k=0}^{5} \dbinom{5}{k}
= \Sum*{  \binomial{5}{\kyw}  }{\kyw}{0}{5} =2^5 = \fpeval{2^5}$
\end{document}

I know nothing about CAS systems, nor am I sure I performed the the examples correctly, but one possibility is to transfer the LaTeX calculations to a language more appropriate for mathematical computation, such as R or Python, and let LaTeX handle the formatting through Quarto markdown. I haven't tested it in this case, but it should faster, and IMHO easier to read and edit if you are used to these languages.

If you want to see what LaTeX code is produced to render the images showed, just change format: pdf to format: latex in the header or add keep-tex: true.

You can also explore how to insert the calculations directly into a pure LaTeX code using knitr or Sagetex, but indeed is easier with Quarto.

---
format: pdf
execute:
  echo: false
documentclass: standalone
classoption: [margin=5pt, varwidth]
pdf-engine: lualatex
---

With **R** and Lua\LaTeX:  \bigskip


```{r, cuentas, eval=FALSE}
start <- min(k) 
end <- max(k)
sumint <- sum(k)
```

```{r}
k <- 1:100
```
```{r, cuentas, eval=TRUE}
```

Test 1: \boxed{\sum_{k=`{r} start`}^{`{r} end`} k = `{r} sumint`}\par\bigskip

```{r}
k <- 0:5
```
```{r, cuentas, eval=TRUE}
```

Test 2: \boxed{\sum_{k=`{r} start`}^{`{r} end`} \binom{`{r} end`}{k} = `{r} sum(choose(max(k),k))` = 2^{`{r} end`} = `{r} 2^end`}\par\bigskip


```{r}
k <- 7:13
```
```{r, cuentas, eval=TRUE}
```

Test 3: \boxed{\sum_{k=`{r} start`}^{`{r} end`} \binom{`{r} end`}{k} = `{r} sum(choose(max(k),k))` = 2^{`{r} end-1`} = `{r} 2^(end-1)`}\par\bigskip

---
format: pdf
execute:
  echo: false
# jupyter: python3 # by defaul if python first
documentclass: standalone
classoption: [margin=5pt, varwidth]
pdf-engine: lualatex
---

With Python and Lua\LaTeX:  \bigskip


```{python}
import math
```

```{python}
start = 1
end = 100
sum_int = sum(range(start, end + 1))
```

Test 1:  \boxed{\sum_{k=`{python} start` }^{`{python} end`} k = `{python} sum_int`}\par\bigskip 


```{python}
start = 0
end = 5
sum_binomial = sum(math.comb(end, k) for k in range(end+1))
potencia = 2**end
```

Test 2: \boxed{\sum_{k=`{python} start`}^{`{python} end`} \binom{`{python} end`}{k} = `{python} sum_binomial` = 2^{`{python} end`} = `{python} potencia`}\par\bigskip

```{python}
start = 7
end = 13
sum_binomial = sum(math.comb(end, k) for k in range(start, end))
potencia = 2**(end-1)
```

Test 3: \boxed{\sum_{k=`{python} start`}^{`{python} end`} \binom{`{python} end`}{k} = `{python} sum_binomial` = 2^{`{python} end-1`} = `{python} potencia`}\par\bigskip


```{python}
start = 1
end = 10
sum_real = sum(1 / k for k in range(start, end + 1))
```

Test 4: \boxed{\sum_{k=`{python} start`}^{`{python} end`} \frac{1}{k} \approx `{python} round(sum_real, 4)`}