I really can't remember (if I have ever known this):

Talamancoeb 2022-01-05 Answered
I really can't remember (if I have ever known this): which series is this and how to demonstrate its solution?
\(\displaystyle{\sum_{{{i}={1}}}^{{n}}}{i}={\frac{{{n}{\left({n}+{1}\right)}}}{{{2}}}}\)

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Expert Answer

Anzante2m
Answered 2022-01-06 Author has 280 answers
Write out this sum twice, once is direct order, and once in reverse:
\(\displaystyle{1}+{2}+\ldots+{\left({n}-{1}\right)}+{n}={s}\)
\(\displaystyle{n}+{\left({n}-{1}\right)}+\ldots+{2}+{1}={s}\)
Now add up column-wise:
\(\displaystyle{\left({n}+{1}\right)}+{\left({n}+{1}\right)}+\ldots+{\left({n}+{1}\right)}+{\left({n}+{1}\right)}={2}{s}\)
There are exactly n terms here (as many as the number of terms in the sum). Hence:
\(\displaystyle{n}{\left({n}+{1}\right)}={2}{s}\)
Now solve for s.
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Jonathan Burroughs
Answered 2022-01-07 Author has 2167 answers
This is an Arithmetic Series starting from 1 with difference 1.
\(\displaystyle{\sum_{{{i}={1}}}^{{n}}}{i}={1}+{2}+{3}+{4}+\ldots+{n}={\frac{{{n}{\left({n}+{1}\right)}}}{{{2}}}}\)
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karton
Answered 2022-01-11 Author has 8659 answers

This is not a series. This sum is named Gauss sum and that formula \(\frac{n(n+1)}{2}\) you can prove it using induction.
The exercise starts from the following sum: 1+2+...+100 and the way you can classify the terms of this sum.
1+2+...+100=(1+100)+(2+99)+...(50+51)

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