If you have the sums $ (1+2+.. +n) + (1+2+3+.. +n-1)+ (1+2+3+.. +n-2)+(1+2+3+.. +n-3)+... +(1+2+3)+(1+2)+1$for large enough $n$
$$\frac {n^3}{3!} \approx (1+2+.. +n) + (1+2+3+.. +n-1)+ (1+2+3+.. +n-2)+(1+2+3+.. +n-3)+... +(1+2+3)+(1+2)+1$$ if divided the sum by the divisor let's call it $x$ (can be any number $1,2,3.. $) we get $$\frac {n^3}{3!x^2} \approx (1x+2x+.. +\frac {n}{x}x) + (1x+2x+3x+.. +\frac {n-1}{x}x)+ (1x+2x+3x+.. +\frac {n-2}{x}x)+(1x+2x+3x+.. +\frac {n-3}{x}x)+... +(1x+2x+3x)+(1x+2x)+1x$$If the difference between the closest numbers let's call it $d$, $d=\frac {1}{10^k}$, we get $$\frac {n^3}{3!x^2d^2} \approx (1dx+2dx+3dx.. +\frac {n}{x}x) + (1dx+2dx+3dx+.. +\frac {n-d}{x}x)+ (1dx+2dx+3dx+.. +\frac {n-2d}{x}x)+(1dx+2dx+3dx.. +\frac {n-3d}{x}x)+... +(1dx+2dx+3dx)+(1dx+2dx)+1dx$$ if we assume $k\to\infty$ we get $$\frac {n^3}{3!x^2d^2} = (1dx+2dx+3dx.. +\frac {n}{x}x) + (1dx+2dx+3dx+.. +\frac {n-d}{x}x)+ (1dx+2dx+3dx+.. +\frac {n-2d}{x}x)+(1dx+2dx+3dx.. +\frac {n-3d}{x}x)+... +(1dx+2dx+3dx)+(1dx+2dx)+1dx$$ OR $$\frac {n^3}{3!x^2} = (1d^3x+2d^3x+3d^3x.. +\frac {n}{x}dx) + (1d^3x+2d^3x+3d^3x+.. +\frac {n-d}{x}d^2x)+ (1d^3x+2d^3x+3d^3x+.. +\frac {n-2d}{x}d^2x)+(1d^3x+2d^3x+3d^3x.. +\frac {n-3d}{x}d^2x)+... +(1d^3x+2d^3x+3d^3x)+(1d^3x+2d^3x)+1d^3x$$ Now if we assume that $n$ value of circle arc and $x$ value of diameter, we get $n - \frac {n^3}{3!x^2} \approx$ value of chord, So we got the first condition of Taylor series.The question is my calculations are correct?

Unfortunately the LaTeX editor on this system requires the use of [ and delineators, not dollar signs. I am rewriting your question here using those delineators so that it can be more easily understood:

If you have the sums

\normalsize (1+2+.. +n) + (1+2+3+.. +n-1)+ (1+2+3+.. +n-2)+(1+2+3+.. +n-3)+... +(1+2+3)+(1+2)+1

for large enough n

\normalsize \frac {n^3}{3!} \approx (1+2+.. +n) + (1+2+3+.. +n-1)+ (1+2+3+.. +n-2)+(1+2+3+.. +n-3)+... +(1+2+3)+(1+2)+1

if divided the sum by the divisor let's call it x (can be any number 1,2,3.. ) we get

\normalsize \frac {n^3}{3!x^2} \approx (1x+2x+.. +\frac {n}{x}x) + (1x+2x+3x+.. +\frac {n-1}{x}x)+ (1x+2x+3x+.. +\frac {n-2}{x}x)+(1x+2x+3x+.. +\frac {n-3}{x}x)+... +(1x+2x+3x)+(1x+2x)+1x

If the difference between the closest numbers let's call it d, d=\frac {1}{10^k}, we get

\normalsize \frac {n^3}{3!x^2d^2} \approx (1dx+2dx+3dx.. +\frac {n}{x}x) + (1dx+2dx+3dx+.. +\frac {n-d}{x}x)+ (1dx+2dx+3dx+.. +\frac {n-2d}{x}x)+(1dx+2dx+3dx.. +\frac {n-3d}{x}x)+... +(1dx+2dx+3dx)+(1dx+2dx)+1dx

if we assume k\to\infty we get

\normalsize \frac {n^3}{3!x^2d^2}= (1dx+2dx+3dx.. +\frac {n}{x}x) + (1dx+2dx+3dx+.. +\frac {n-d}{x}x)+ (1dx+2dx+3dx+.. +\frac {n-2d}{x}x)+(1dx+2dx+3dx.. +\frac {n-3d}{x}x)+... +(1dx+2dx+3dx)+(1dx+2dx)+1dx

OR

\normalsize \frac {n^3}{3!x^2} = (1d^3x+2d^3x+3d^3x.. +\frac {n}{x}dx) + (1d^3x+2d^3x+3d^3x+.. +\frac {n-d}{x}d^2x)+ (1d^3x+2d^3x+3d^3x+.. +\frac {n-2d}{x}d^2x)+(1d^3x+2d^3x+3d^3x.. +\frac {n-3d}{x}d^2x)+... +(1d^3x+2d^3x+3d^3x)+(1d^3x+2d^3x)+1d^3x

Now if we assume that n value of circle arc and x value of diameter, we get \small n - \frac {n^3}{3!x^2} \approx. value of chord, So we got the first condition of Taylor series.The question is my calculations are correct?
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Now for my comment on your post: starting with the line "if divided the sum by the divisor let's call it x" you have multiple math errors, which makes it difficult to understand what you are trying to do. I suggest you check your LaTeX nomenclature, correct the errors, and repost using the proper [math ] and [ /math] delineators for LaTeX.