Question #78b00
1 Answer
Jul 6, 2017
sum_(r=0)^4 m(3r-6) 4∑r=0m(3r−6)
Explanation:
We have a sequence of terms:
-6m, - 3m, 0, 3m, 6m −6m,−3m,0,3m,6m
The difference between the first and second term is:
-3m -( - 6m) = 3m −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 −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 un=−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)
