Sigma notation - Mathcentre

Solution
Notice that, in this example, there are 6 terms in the sum, because we have k = 0 for the first
term:
5∑
2 k = 2 0 + 2 1 + 2 2 + 2 3 + 2 4 + 2 5
Example
6∑
1
Evaluate r(r + 1).
2
Solution
r=1
k=0
= 1 + 2 + 4 + 8 + 16 + 32
= 63 .
You might recognise that each number 1 r(r + 1) is a triangular number, and so this example
2
asks for the sum of the first six triangular numbers. We get
6∑
r=1
1
2 r(r + 1) = ( 1
2 × 1 × 2) + ( 1
2 × 2 × 3) + ( 1
2 × 3 × 4) + ( 1
2 × 4 × 5)
+ ( 1
2 × 5 × 6) + ( 1
2 × 6 × 7)
= 1 + 3 + 6 + 10 + 15 + 21
= 56 .
What would we do if we were asked to evaluate
n∑
2 k ?
k=1
Now we know what this expression means, because it is the sum of all the terms 2 k where k
takes the values from 1 to n, and so it is
n∑
2 k = 2 1 + 2 2 + 2 3 + 2 4 + . . . + 2 n .
k=1
But we cannot give a numerical answer, as we do not know the value of the upper limit n.
Example
4∑
Evaluate (−1) r .
Solution
r=1
Here, we need to remember that (−1) 2 = +1, (−1) 3 = −1, and so on. So
4∑
(−1) r = (−1) 1 + (−1) 2 + (−1) 3 + (−1) 4
r=1
= (−1) + 1 + (−1) + 1
= 0 .
c○ mathcentre July 18, 2005 4