# How do you find the partial sum of Sigma n from n=1 to 50?

Dec 3, 2017

${\sum}_{n = 1}^{50} n = 1275$

#### Explanation:

We can evaluate the sum the tedious way by just adding continually, but we can be smart. There is a formula for adding all the numbers up to $k$:

${\sum}_{n = 1}^{k} n = \frac{k \left(k + 1\right)}{2}$

In this case, it becomes:
${\sum}_{n = 1}^{50} n = \frac{50 \left(50 + 1\right)}{2} = \frac{50 \cdot 51}{2} = 1275$

Dec 3, 2017

${\Sigma}_{n = 1}^{50} n = \textcolor{b l u e}{1275}$

#### Explanation:

s=(Sigma_(n=1)^50 n) = overbrace(color(white)("x")1+ ... + 50)^(50" terms")
$s = \left({\Sigma}_{n = 1}^{50} n\right) = {\underbrace{50 + \ldots + 1}}_{50 \text{ terms}}$

$2 s \textcolor{w h i t e}{\text{xxxxxxxxxx")=underbrace(51+ ... +51)_(50" terms}}$

$\textcolor{w h i t e}{\text{xxxxxxxxxxxxxx}} = 51 \times 50$

$\rightarrow s = \left({\Sigma}_{n = 1}^{50} n\right) = 51 \times 25 = 1275$