Question #78b00

Question

1661 views around the world

You can reuse this answer
Creative Commons License

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:

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