# How do you find the partial sum of Sigma 2n from n=1 to 100?

Feb 20, 2018

${\sum}_{n = 1}^{100} \setminus 2 n = 10100$

#### Explanation:

We seek the sum:

${S}_{100} = {\sum}_{n = 1}^{100} \setminus 2 n$

We can readily find the general sum:

${S}_{n} = {\sum}_{r = 1}^{n} \setminus 2 r$
$\setminus \setminus \setminus \setminus = 2 \setminus {\sum}_{r = 1}^{n} \setminus r$

And. we use the standard formula for the sum of the first $n$ integers:

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

Giving us:

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

So with $n = 100$ we have:

${S}_{100} = 100 \left(101\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 10100$