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

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

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Anzante2m
Write out this sum twice, once is direct order, and once in reverse:
$$\displaystyle{1}+{2}+\ldots+{\left({n}-{1}\right)}+{n}={s}$$
$$\displaystyle{n}+{\left({n}-{1}\right)}+\ldots+{2}+{1}={s}$$
$$\displaystyle{\left({n}+{1}\right)}+{\left({n}+{1}\right)}+\ldots+{\left({n}+{1}\right)}+{\left({n}+{1}\right)}={2}{s}$$
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.
###### Not exactly what youâ€™re looking for?
Jonathan Burroughs
This is an Arithmetic Series starting from 1 with difference 1.
$$\displaystyle{\sum_{{{i}={1}}}^{{n}}}{i}={1}+{2}+{3}+{4}+\ldots+{n}={\frac{{{n}{\left({n}+{1}\right)}}}{{{2}}}}$$
karton

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)