# How do you evaluate \sum _ { n = 1} ^ { 20} ( n + n ^ { 3} ) ?

May 10, 2017

${\sum}_{n = 1}^{20} \left(n + {n}^{3}\right) = 44310$

#### Explanation:

Let:

$S = {\sum}_{n = 1}^{20} \left(n + {n}^{3}\right)$

We need the standard formula:

${\sum}_{r = 1}^{n} r \setminus = \frac{1}{2} n \left(n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$

Which gives us:
$S = {\sum}_{n = 1}^{20} n + {\sum}_{n = 1}^{20} {n}^{3}$
$\setminus \setminus = \frac{1}{2} \left(20\right) \left(21\right) + \frac{1}{4} \left({20}^{2}\right) \left({21}^{2}\right)$
$\setminus \setminus = \frac{1}{2} \left(420\right) + \frac{1}{4} \left(400\right) \left(441\right)$
$\setminus \setminus = 210 + \left(100\right) \left(441\right)$
$\setminus \setminus = 210 + 44100$
$\setminus \setminus = 44310$