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I am very sorry since this is the first time I am here and I unsure how to type any math.

Anyway.

The sum of all numbers to any n is (n^2+n)/2, and I divided that by n to get the average at that point.

And, the sum of all numbers at n-1 ((n-1)^2+n-1)/2, and I divided that by n-1 to the get the average at n-1. Then I just added + n, and divided by 2. However, I am not getting the same answer as just taking the average of the sum of all the numbers to a number n.

Hope my question made sense, and someone can explain why the 2 are not equal.

Indraditya Sarkar
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    You need to put $ before and after your LaTeX to get it to display properly – Henry Jun 26 '22 at 00:33
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    Welcome to SE! You’ll need to put $ before and after your TeX code. Also it’s unclear what you’re asking about so you’ll have to put more context into your questions. – blakedylanmusic Jun 26 '22 at 00:33
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    $\dfrac{\left(\frac{\left(n^{2}+n\right)}{2}\right)}{n} = \dfrac{n+1}{2}$ which is the average of the last and first terms of the sequence, as well as the average of all the terms in $1,2,\ldots,n$ – Henry Jun 26 '22 at 00:34
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    If you are asking why $;\dfrac{1+2+3}{3} \ne \dfrac{\dfrac{1+2}{2}+3}{2};$ then they are not equal, indeed. Averages can be combined, but you would need to use weighted averages instead e.g. $;\dfrac{1+2+3}{3} = \dfrac{\color{red}{2 \cdot}\dfrac{1+2}{2}+\color{red}{1 \cdot}3}{\color{red}{2+1}},$. – dxiv Jun 26 '22 at 00:35
  • @Henry wouldn't ((n^2+n)/2)/n= (n^3+n^2)/2 – Indraditya Sarkar Jun 26 '22 at 00:39
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    @IndradityaSarkar You still need $ before and after using $((n^2+n)/2)/n= (n^3+n^2)/2$ would give $((n^2+n)/2)/n= (n^3+n^2)/2$ that that is incorrect. In fact $\dfrac{\left(\frac{\left(n^{2}+n\right)}{2}\right)}{n} = \dfrac{\left(n^{2}+n\right)}{2n}= \dfrac{\left(n+1\right)}{2}$ Try it with $n=3$ or something else – Henry Jun 26 '22 at 00:43
  • @IndradityaSarkar You're right, $((n^2+n)/2)/n$ wouldn't $= (n^3+n^2)/2$ – peterwhy Jun 26 '22 at 00:44
  • Why should you get the same answer? You get an average by adding $n$ numbers together and dividing by $n$. You aren't doing that at all. You are adding two number and dividing by $2$. That's an average of $2$ numbers; not $n$. But you can fix it. You have the average of the $n-1$ numbers. Take that average and multiply by $n-1$ to get the total sum and then you add an $n$th number to it and divide by $n$. That will work. – fleablood Jun 26 '22 at 00:54

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