Question #78b00

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Answer 1 · 2017-07-06T22:03:58

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

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 #r^(th)# term of the sequence by:

# 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) #