# What is the sum of the arithmetic series 1,2,3,4...80?

It is an arithmetic series with first term ${a}_{1} = 1$ and ratio $r = 1$
hence

${S}_{80} = \left(\frac{80}{2}\right) \cdot \left(1 + 80\right) = 3240$

Feb 11, 2016

$3240$

#### Explanation:

The sum of a finite arithmetic sequence is equal to the number of terms multiplied by the average term. The average term is the same as the average of the first and last term.

So in our example:

${\sum}_{n = 1}^{80} n = 80 \cdot \frac{1 + 80}{2} = 40 \cdot 81 = 3240$

Feb 11, 2016

$3240$

#### Explanation:

A good way to envision how to do this is to imagine pairs:

Start with the largest and smallest terms of the sequence: $80$ and $1$.

$80 + 1 = 81$

The next largest and smallest are $79$ and $2$, which have the same sum.

$79 + 2 = 81$

We can start listing these pairs:

$80 + 1 = 81$
$79 + 2 = 81$
$78 + 3 = 81$
$77 + 4 = 81$
$76 + 5 = 81$

$\ldots \text{continue} \ldots$

$43 + 38 = 81$
$42 + 39 = 81$
$41 + 40 = 81$

In total, there are $40$ pairs since $\frac{80}{2} = 40$.

$40$ pairs of $81$ each is $40 \times 81 = \textcolor{red}{3240}$.