# Question #78b00

##### 1 Answer

Jul 6, 2017

#### Answer:

# sum_(r=0)^4 m(3r-6) #

#### Explanation:

We have a sequence of terms:

# -6m, - 3m, 0, 3m, 6m #

The difference between the first and second term is:

# -3m -( - 6m) = 3m #

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:

# -6m+0*3m, -6m+1*3m, -6m+2*3m, -6m+3*3m, -6m+4*3m #

Thus we can denote the

# u_n = -6m+r*3m #

# \ \ \ \ = m(3r-6) #

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

# sum_(r=0)^4 m(3r-6) #