# Question #78b00

Jul 6, 2017

#### Answer:

${\sum}_{r = 0}^{4} m \left(3 r - 6\right)$

#### Explanation:

We have a sequence of terms:

$- 6 m , - 3 m , 0 , 3 m , 6 m$

The difference between the first and second term is:

$- 3 m - \left(- 6 m\right) = 3 m$

And is a trivial exercise to show that this is also the difference between further consecutive terms. So we can write the series as follows:

$- 6 m + 0 \cdot 3 m , - 6 m + 1 \cdot 3 m , - 6 m + 2 \cdot 3 m , - 6 m + 3 \cdot 3 m , - 6 m + 4 \cdot 3 m$

Thus we can denote the ${r}^{t h}$ term of the sequence by:

${u}_{n} = - 6 m + r \cdot 3 m$
$\setminus \setminus \setminus \setminus = m \left(3 r - 6\right)$

Hence, we can denote sum of this sequence using sigma notation as follows:

${\sum}_{r = 0}^{4} m \left(3 r - 6\right)$