# How do you use the properties of summation to evaluate the sum of Sigma (i^3-2i) from i=1 to 15?

Dec 24, 2016

The answer is $= 14160$

#### Explanation:

We need

${\sum}_{1}^{n} {r}^{3} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$

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

Therefore,

${\sum}_{1}^{15} \left({i}^{3} - 2 i\right)$

$= {\sum}_{1}^{15} {i}^{3} - 2 {\sum}_{1}^{15} i$

$= \frac{1}{4} \cdot {15}^{2} \left({16}^{2}\right) - 2 \cdot \frac{1}{2} \cdot 15 \cdot 16$

$= 15 \cdot 16 \left(15 \cdot \frac{16}{4} - 1\right)$

$= 15 \cdot 16 \left(60 - 1\right)$

$= 15 \cdot 16 \cdot 59$

$= 14160$