Summation: How to Solve
How to solve the given summation by using the summation properties: 3 examples and their solutions.
Example 1
Example
Solution
See the first term 5ak.
5 is a coefficient.
So 5 gets out from the sigma.
The variable of the sigma is k.
So ak is in the sigma.
This is true because
∑ 5ak
= 5a1 + 5a2 + 5a3 + ... + 5an
= 5(a1 + a2 + a3 + ... + an)
= 5 ∑ ak.
Sigma Notation
Next, see the next term 2bk.
+2 is a coefficient.
So +2 gets out from the sigma.
The variable of the sigma is k.
So bk is in the sigma.
So ∑ (5ak + 2bk)
= 5 ∑ ak + 2 ∑ bk.
∑ ak = 7
∑ bk = -3
So 5 ∑ ak + 2 ∑ bk
= 5⋅7 + 2⋅(-3).
5⋅7 = 35
+2⋅(-3) = -6
35 - 6 = 29
So 29 is the answer.
Example 2
Example
Solution
See the first term 2ak.
2 is a coefficient.
So 2 gets out from the sigma.
The variable of the sigma is k.
So ak is in the sigma.
Next, see -7.
-7 is a constant.
Then it becomes -7n.
This is true because
∑ (-7) means
adding -7 n times: -7n.
So ∑ (2ak -7)
= 2 ∑ ak - 7n.
∑ ak = 5
So 2 ∑ ak - 7n = 2⋅5 - 7n.
2⋅5 = 10
Arrange the terms in descending order.
So -7n + 10 is the answer.
Example 3
Example
Solution
See the term nak.
The variable of the sigma is k, not n.
So think n as a constant, in this case.
Then n gets out from the sigma.
And ak is in the sigma.
∑ ak = n2 + 1
So n ∑ ak = n(n2 + 1).
n(n2 + 1) = n3 + n
So n3 + n is the answer.